Dirac Delta Functions - Differential Equations

1. What is the Laplace transform of $g(t) = t^2 \delta(t-4)$?
   Answer: $16e^{-4s}$ When you encounter a Laplace transform involving the Dirac delta function $\delta(t-a)$, remember that this function has a special "sifting" property that simplifies calculations significantly. To find $\mathcal{L}\{t^2 \delta(t-4)\}$, you need to use the key property of the delta function: $\int_{0}^{\infty} f(t)\delta(t-a) e^{-st} dt = f(a)e^{-as}$ for $a > 0$. This means the delta function "picks out" the value of $f(t)$ at $t = a$. Here, $f(t) = t^2$ and $a = 4$, so the delta function evaluates $t^2$ at $t = 4$, giving us $f(4) = 4^2 = 16$. The exponential term becomes $e^{-4s}$. Therefore: $\mathcal{L}\{t^2 \delta(t-4)\} = 16e^{-4s}$ Let's examine why the other answers miss the mark. Choice A ($e^{-4s}$) forgot to evaluate $t^2$ at $t = 4$ and only applied the shifting property. Choice B ($e^{-4s} \frac{2}{s^3}$) incorrectly tried to use the standard Laplace transform of $t^2$ (which is $\frac{2}{s^3}$) while applying the shifting theorem, but this approach doesn't work with delta functions. Choice C ($16$) correctly evaluated $t^2$ at $t = 4$ but forgot the exponential shifting factor $e^{-4s}$. Study tip: When you see delta functions in Laplace transforms, ignore the standard transform formulas for polynomials. Instead, use the sifting property: evaluate the non-delta part at the shift point and multiply by $e^{-as}$.
2. Consider the function $f(t) = 3\delta(t-2) + 5\delta(t-4)$. If $g(t) = \int_{0}^{t} f(\tau) d\tau$, what is the value of $g(5)$?
   Answer: $g(5) = 8$ The correct answer is A. Since $g(t) = \int_{0}^{t} f(\tau) d\tau = \int_{0}^{t} [3\delta(\tau-2) + 5\delta(\tau-4)] d\tau$, we evaluate this integral by using the property that $\int_{a}^{b} \delta(t-c) dt = 1$ if $a < c < b$, and 0 otherwise. For $g(5)$, the interval is $[0,5]$, which contains both impulses at $t=2$ and $t=4$. Therefore, $g(5) = 3 \cdot 1 + 5 \cdot 1 = 8$. Choice B incorrectly includes only the second impulse. Choice C incorrectly includes only the first impulse. Choice D incorrectly multiplies by the impulse locations.
3. Consider the solution $y(t)$ to the initial value problem $y'' + p(t)y' + q(t)y = A\delta(t-t_0)$ with $y(0)=y'(0)=0$, where $p(t)$ and $q(t)$ are continuous functions and $t_0 > 0$. Which of the following statements about the solution at $t=t_0$ must be true?
   Answer: $y(t_0) = 0$ and $y'(t)$ is discontinuous at $t=t_0$. When you encounter a differential equation with a Dirac delta function $\delta(t-t_0)$ as forcing term, you're dealing with an impulse response problem. The key insight is understanding how delta functions affect continuity of solutions and their derivatives. The delta function represents an instantaneous impulse at $t = t_0$. For the equation $y'' + p(t)y' + q(t)y = A\delta(t-t_0)$, this impulse creates a sudden jump in the first derivative while keeping the solution itself continuous. Here's why: integrating the differential equation across the impulse from $t_0^-$ to $t_0^+$, the delta function contributes $A$ to the integral, while the $y'$ and $y$ terms (being bounded) contribute zero over the infinitesimal interval. This means $y'(t_0^+) - y'(t_0^-) = A$, creating a jump discontinuity in $y'$. However, $y(t)$ itself remains continuous since $y(t_0^+) - y(t_0^-) = 0$. Given the initial conditions $y(0) = y'(0) = 0$ and that the impulse occurs at $t_0 > 0$, the solution remains zero until the impulse: $y(t_0) = 0$. Choice A incorrectly suggests $y$ jumps to $A$. Choice C wrongly claims $y'(t_0) = A$ when actually $y'$ has a jump of magnitude $A$, not a specific value. Choice D incorrectly states that $y$ itself is discontinuous. Remember: delta function impulses create jump discontinuities in derivatives, not in the solution itself. The magnitude of the jump equals the coefficient of the delta function.
4. Find the solution to the initial value problem $y'' + \omega^2 y = \delta(t)$, with initial conditions $y(0)=0$ and $y'(0)=0$.
   Answer: $\frac{1}{\omega}\sin(\omega t)$ This problem involves solving a differential equation with a Dirac delta function, which represents an instantaneous impulse. When you see $\delta(t)$ on the right side of a differential equation, think about using the Laplace transform method, as it handles discontinuous functions elegantly. Taking the Laplace transform of $y'' + \omega^2 y = \delta(t)$ with zero initial conditions gives us $s^2Y(s) + \omega^2Y(s) = 1$, where $Y(s) = \mathcal{L}\{y(t)\}$. The Laplace transform of $\delta(t)$ is simply 1. Solving for $Y(s)$: $Y(s) = \frac{1}{s^2 + \omega^2}$. Taking the inverse Laplace transform, we get $y(t) = \frac{1}{\omega}\sin(\omega t)$, which is answer C. Let's examine why the other options are incorrect. Option A, $\frac{1}{\omega^2}(1-\cos(\omega t))$, would be the solution if the forcing function were a unit step function rather than a delta function. Option B, $\sin(\omega t)$, is missing the crucial $\frac{1}{\omega}$ factor that comes from the inverse Laplace transform. Option D suggests no response occurs, which contradicts the physical reality that an impulse input produces a response in any second-order system. Remember that delta function problems often involve impulse responses, and the Laplace transform is your best tool. The key insight is that $\mathcal{L}^{-1}\left\{\frac{1}{s^2 + \omega^2}\right\} = \frac{1}{\omega}\sin(\omega t)$, not just $\sin(\omega t)$. Always include scaling factors when applying inverse transforms.
5. A mass-spring system is described by the differential equation $y'' + 4y = \delta(t-\pi) + \delta(t-2\pi)$, with initial conditions $y(0) = 0$ and $y'(0) = 0$. What is the value of $y(3\pi)$?
   Answer: $y(3\pi) = \frac{1}{2}\sin(4\pi) + \frac{1}{2}\sin(2\pi) = 0$ The correct answer is B. Taking the Laplace transform: $s^2Y(s) + 4Y(s) = e^{-\pi s} + e^{-2\pi s}$, so $Y(s) = \frac{e^{-\pi s} + e^{-2\pi s}}{s^2 + 4}$. Using $\mathcal{L}^{-1}\{\frac{e^{-as}}{s^2 + 4}\} = \frac{1}{2}u(t-a)\sin(2(t-a))$, we get $y(t) = \frac{1}{2}u(t-\pi)\sin(2(t-\pi)) + \frac{1}{2}u(t-2\pi)\sin(2(t-2\pi))$. At $t = 3\pi$: $y(3\pi) = \frac{1}{2}\sin(2(3\pi-\pi)) + \frac{1}{2}\sin(2(3\pi-2\pi)) = \frac{1}{2}\sin(4\pi) + \frac{1}{2}\sin(2\pi) = 0$. Choices A and C have computational errors in the argument evaluation. Choice D incorrectly computes the sine arguments.
6. Consider the initial value problem $y'' + 2y' + 5y = \delta(t-c)$, where $c > 0$, and $y(0)=y'(0)=0$. Let $y(t)$ be the solution. What is the value of the jump in the derivative at $t=c$, defined as $\lim_{t \to c^+} y'(t) - \lim_{t \to c^-} y'(t)$?
   Answer: 1 For a linear second-order differential equation of the form $ay'' + by' + cy = \delta(t-c)$, the Dirac delta function on the right-hand side causes a jump discontinuity in the first derivative $y'(t)$ at $t=c$. The magnitude of this jump is $1/a$. In this problem, the equation is $y'' + 2y' + 5y = \delta(t-c)$, so the coefficient $a$ of $y''$ is 1. Therefore, the jump in the derivative is $\lim_{t \to c^+} y'(t) - \lim_{t \to c^-} y'(t) = 1/1 = 1$. The other terms in the equation and the location $c$ of the impulse do not affect the magnitude of the jump.
7. A delta function $\delta(t-a)$ can be approximated by a sequence of rectangular pulses. If $\delta_n(t-a) = \frac{n}{2}$ for $a - \frac{1}{n} \leq t \leq a + \frac{1}{n}$ and $\delta_n(t-a) = 0$ elsewhere, what happens to $\int_{-\infty}^{\infty} f(t)\delta_n(t-a) dt$ as $n \to \infty$ for a continuous function $f(t)$?
   Answer: The integral approaches $f(a)$ due to the sifting property of delta functions This question tests your understanding of how delta function approximations work and the fundamental sifting property that emerges in the limit. Let's analyze what happens as $n \to \infty$. The rectangular pulse $\delta_n(t-a)$ has height $\frac{n}{2}$ and width $\frac{2}{n}$ (from $a-\frac{1}{n}$ to $a+\frac{1}{n}$). Notice that the area under each pulse is $\frac{n}{2} \cdot \frac{2}{n} = 1$, which remains constant for all $n$. As $n$ increases, the pulse becomes infinitely tall and infinitesimally narrow while maintaining unit area. This is precisely how we construct the Dirac delta function. The integral becomes: $\int_{-\infty}^{\infty} f(t)\delta_n(t-a) dt = \int_{a-\frac{1}{n}}^{a+\frac{1}{n}} f(t) \cdot \frac{n}{2} dt$ Since $f(t)$ is continuous, as the interval shrinks around $t = a$, we can factor out $f(a)$ (by continuity), giving us approximately $f(a) \cdot \frac{n}{2} \cdot \frac{2}{n} = f(a)$. In the limit, this becomes exactly $f(a)$ — the sifting property. Choice A incorrectly suggests we get the derivative; the shrinking width alone doesn't create differentiation. Choice B gets the wrong factor because it ignores that both height increases and width decreases proportionally. Choice D makes an error with the factor of 2, misunderstanding how the height and width relationship works. Remember: Delta function approximations always maintain unit area, and the sifting property $\int f(t)\delta(t-a)dt = f(a)$ is the defining characteristic you should expect in any limiting process.
8. Which of the following expressions represents the derivative of the Heaviside step function $u(t-a)$ in terms of the Dirac delta function?
   Answer: $\frac{d}{dt}u(t-a) = \delta(t-a)$ The correct answer is B. The derivative of the Heaviside step function $u(t-a)$ is the Dirac delta function $\delta(t-a)$. This fundamental relationship shows that the delta function represents an instantaneous unit impulse at $t=a$, which is exactly what happens when the step function jumps from 0 to 1. Choice A incorrectly suggests a difference of delta functions. Choice C incorrectly multiplies the step function with the delta function, which would always be zero since they don't overlap. Choice D has the wrong sign in the argument of the delta function.
9. Consider the differential equation $y' + 2y = 3\delta(t-1) + \delta(t-3)$ with $y(0) = 1$. What is the value of $\lim_{t \to 1^+} y(t) - \lim_{t \to 1^-} y(t)$?
   Answer: The jump discontinuity is equal to $3$ When you encounter differential equations with delta functions (impulses), focus on how these instantaneous forces create jump discontinuities in the solution. The delta function $\delta(t-a)$ represents an impulse at $t = a$, and the coefficient tells you the magnitude of the jump. For this first-order linear equation $y' + 2y = 3\delta(t-1) + \delta(t-3)$, the solution will have jump discontinuities at $t = 1$ and $t = 3$. The key insight is that when an impulse $c\delta(t-a)$ appears on the right side of $y' + p(t)y = f(t)$, it causes the solution to jump by exactly $c$ at $t = a$. At $t = 1$, we have the impulse $3\delta(t-1)$, so the solution jumps upward by $3$. This means $\lim_{t \to 1^+} y(t) - \lim_{t \to 1^-} y(t) = 3$. Option A is wrong because delta functions always create jump discontinuities in first-order equations. Option B suggests the jump is $\frac{3}{2}$, which might come from incorrectly dividing the impulse strength by the coefficient of $y$, but this is not how impulse responses work in first-order linear equations. Option C gives a jump of $1$, which would correspond to an impulse $\delta(t-1)$ rather than $3\delta(t-1)$. Remember: in first-order linear differential equations, an impulse $c\delta(t-a)$ creates a jump discontinuity of magnitude $c$ at $t = a$. The jump magnitude equals the impulse coefficient directly.
10. The function $p(t) = \delta(t-2) - \delta(t-4)$ represents the difference of two impulses. What is $\mathcal{L}^{-1}\left\{\frac{P(s)}{s+3}\right\}$ where $P(s) = \mathcal{L}\{p(t)\}$?
    Answer: $\mathcal{L}^{-1}\left\{\frac{P(s)}{s+3}\right\} = e^{-3(t-2)}u(t-2) - e^{-3(t-4)}u(t-4)$ When you encounter problems involving delta functions and Laplace transforms, you need to systematically apply the properties of impulse functions and the convolution theorem (or equivalently, the multiplication property in the s-domain). First, find the Laplace transform of $p(t) = \delta(t-2) - \delta(t-4)$. Using the shifting property of delta functions: $P(s) = e^{-2s} - e^{-4s}$. Now you need $\mathcal{L}^{-1}\left\{\frac{P(s)}{s+3}\right\} = \mathcal{L}^{-1}\left\{\frac{e^{-2s} - e^{-4s}}{s+3}\right\}$. This splits into two terms: $\mathcal{L}^{-1}\left\{\frac{e^{-2s}}{s+3}\right\} - \mathcal{L}^{-1}\left\{\frac{e^{-4s}}{s+3}\right\}$. Since $\mathcal{L}^{-1}\left\{\frac{1}{s+3}\right\} = e^{-3t}$, applying the second shifting theorem gives us: $\mathcal{L}^{-1}\left\{\frac{e^{-as}}{s+3}\right\} = e^{-3(t-a)}u(t-a)$. Therefore: $\mathcal{L}^{-1}\left\{\frac{P(s)}{s+3}\right\} = e^{-3(t-2)}u(t-2) - e^{-3(t-4)}u(t-4)$, which is answer C. Answer A incorrectly factors out $e^{-3t}$ and uses the wrong form for shifted functions. Answer B includes an incorrect factor of $\frac{1}{3}$, likely from confusing integration with the transform process. Answer D incorrectly combines the unit step functions, losing the individual switching behavior at $t=4$. Remember: when you see $\frac{e^{-as}F(s)}{G(s)}$, first find $\mathcal{L}^{-1}\left\{\frac{F(s)}{G(s)}\right\}$, then apply the shifting theorem to get the final result with proper unit step functions.
11. A system receives an input $x(t) = \delta(t) + 2\delta(t-1)$. If the impulse response of the system is $h(t) = e^{-2t}u(t)$, what is the output $y(t)$ for $t \geq 2$?
    Answer: $y(t) = e^{-2t} + 2e^{-2t+2}$ for $t \geq 2$ When you encounter a linear system with impulse response and need to find the output, you're working with convolution. The key insight is that the output equals the convolution of the input with the impulse response: $y(t) = x(t) * h(t)$. Since your input $x(t) = \delta(t) + 2\delta(t-1)$ contains delta functions, you can use the shifting property: convolving with $\delta(t-a)$ shifts the impulse response by $a$ units. This means $y(t) = h(t) + 2h(t-1)$. Substituting $h(t) = e^{-2t}u(t)$: - $h(t) = e^{-2t}u(t)$ - $h(t-1) = e^{-2(t-1)}u(t-1) = e^{-2t+2}u(t-1)$ For $t \geq 2$, both unit step functions $u(t)$ and $u(t-1)$ equal 1, so: $y(t) = e^{-2t} + 2e^{-2t+2}$ This matches answer D. Looking at the wrong answers: A incorrectly writes $e^{-2(t-1)}$ instead of properly expanding the exponent. B attempts to combine terms incorrectly, yielding $3e^{-2t}$ which ignores the time shift in the second impulse. C unnecessarily includes $u(t-1)$ in the final answer even though we're told $t \geq 2$, making this term redundant. Remember: when convolving with shifted delta functions, apply the shifting property directly to the impulse response, then carefully expand any exponential terms. The algebra in the exponents is where most mistakes happen.
12. The Laplace transform of the solution to $y'' + by' + cy = g(t)$ with $y(0)=1, y'(0)=-1$ is $Y(s) = \frac{s-1}{s^2+4s+8} - \frac{3e^{-\pi s}}{s^2+4s+8}$. What is the forcing function $g(t)$?
    Answer: $-4\delta(t) - 3\delta(t-\pi)$ When you encounter a differential equation with its Laplace transform solution already given, you're being asked to work backwards to find the forcing function. The key insight is using the relationship between the Laplace transform of the differential equation and its components. Taking the Laplace transform of $y'' + by' + cy = g(t)$ with initial conditions $y(0)=1, y'(0)=-1$ gives: $s^2Y(s) - sy(0) - y'(0) + b[sY(s) - y(0)] + cY(s) = G(s)$ Substituting the initial conditions: $s^2Y(s) - s + 1 + b[sY(s) - 1] + cY(s) = G(s)$ From the denominator $s^2 + 4s + 8$ in the given solution, we identify $b = 4$ and $c = 8$. This gives us: $(s^2 + 4s + 8)Y(s) - s + 1 - 4 = G(s)$ $(s^2 + 4s + 8)Y(s) - s - 3 = G(s)$ Substituting the given $Y(s) = \frac{s-1}{s^2+4s+8} - \frac{3e^{-\pi s}}{s^2+4s+8}$: $G(s) = (s-1) - 3e^{-\pi s} - s - 3 = -4 - 3e^{-\pi s}$ Since $\mathcal{L}[\delta(t)] = 1$ and $\mathcal{L}[\delta(t-\pi)] = e^{-\pi s}$, we have $g(t) = -4\delta(t) - 3\delta(t-\pi)$. Choice A omits the impulse at $t=0$. Choice B omits both the magnitude and location of the first impulse. Choice D has incorrect signs for both impulses. Strategy tip: When working backwards from Laplace transforms, always use the differential equation structure to set up your algebra systematically, and remember that exponential factors like $e^{-as}$ correspond to time delays in delta functions.
13. The Laplace transform of $h(t) = 2\delta(t-1) \cos(3t)$ is:
    Answer: $\mathcal{L}\{h(t)\} = 2e^{-s}\cos(3)$ The correct answer is B. Using the property $\mathcal{L}\{\delta(t-a)f(t)\} = e^{-as}f(a)$, we have $\mathcal{L}\{2\delta(t-1)\cos(3t)\} = 2e^{-s}\cos(3 \cdot 1) = 2e^{-s}\cos(3)$. Choice A ignores the delta function shift and treats it as a regular cosine transform. Choice C incorrectly attempts to combine the shift theorem with the cosine transform formula. Choice D incorrectly applies the Laplace transform of cosine after the shift, rather than evaluating the cosine at the shift point.
14. A 2-kg mass is attached to a spring with spring constant 8 N/m and a damper with damping coefficient 4 Ns/m. The system is initially at rest in its equilibrium position. At time $t=5$ s, the mass is struck by a hammer, imparting an impulse of 10 Ns. Which of the following initial value problems correctly models the displacement $x(t)$ of the mass?
    Answer: $x'' + 2x' + 4x = 5\delta(t-5)$, $x(0)=0$, $x'(0)=0$ When modeling a mass-spring-damper system with an impulse, you need to set up the differential equation from the physical forces and handle the impulse correctly using the Dirac delta function. The governing equation comes from Newton's second law: $ma = F_{spring} + F_{damper} + F_{external}$. With mass $m = 2$ kg, spring constant $k = 8$ N/m, and damping coefficient $c = 4$ Ns/m, this gives us $2x'' = -8x - 4x' + F_{external}$, or $2x'' + 4x' + 8x = F_{external}$. Dividing by 2 yields $x'' + 2x' + 4x = \frac{F_{external}}{2}$. The impulse of 10 Ns at $t = 5$ s is modeled as $F_{external} = 10\delta(t-5)$, so the forcing term becomes $\frac{10\delta(t-5)}{2} = 5\delta(t-5)$. Since the system starts at rest in equilibrium, $x(0) = 0$ and $x'(0) = 0$. This gives us answer C. Answer A incorrectly ignores the impulse entirely, treating it as an initial velocity condition instead. Answer B uses the wrong coefficient for the delta function—it forgot to divide the impulse by the mass. Answer D makes two errors: it doesn't divide the equation by mass (keeping the original $2x'' + 4x' + 8x$) and uses a unit step function $u(t-5)$ instead of a delta function, which would model a constant force rather than an instantaneous impulse. Remember: impulses are modeled with delta functions, and always divide your equation by the mass coefficient to get standard form before identifying the forcing term.
15. A system is described by the initial value problem $y' + 3y = \delta(t-1) - \delta(t-2)$, with $y(0)=2$. Find the value of $y(3)$.
    Answer: $e^{-6} - e^{-3} + 2e^{-9}$ First, solve the IVP using Laplace transforms. Let $Y(s) = \mathcal{L}\{y(t)\}$. The transformed equation is $sY(s) - y(0) + 3Y(s) = e^{-s} - e^{-2s}$. Substituting $y(0)=2$ gives $(s+3)Y(s) - 2 = e^{-s} - e^{-2s}$. Solving for $Y(s)$, we get $Y(s) = \frac{2}{s+3} + \frac{e^{-s}}{s+3} - \frac{e^{-2s}}{s+3}$. Taking the inverse Laplace transform term by term yields $y(t) = 2e^{-3t} + u(t-1)e^{-3(t-1)} - u(t-2)e^{-3(t-2)}$. To find $y(3)$, we substitute $t=3$. Since $3>1$ and $3>2$, both Heaviside functions $u(3-1)$ and $u(3-2)$ are equal to 1. So, $y(3) = 2e^{-3(3)} + e^{-3(3-1)} - e^{-3(3-2)} = 2e^{-9} + e^{-6} - e^{-3}$.
16. Let $y(t)$ be the solution to $y' + y = \delta(t-2)$ with $y(0)=1$. Evaluate the integral $\int_0^\infty y(t) \delta(t-3) dt$.
    Answer: $e^{-3} + e^{-1}$ This problem combines two key concepts: solving differential equations with impulse functions and evaluating integrals with the Dirac delta function. When you encounter a delta function $\delta(t-a)$ in a differential equation, it represents an instantaneous impulse at $t=a$, causing a jump discontinuity in the solution. To solve $y' + y = \delta(t-2)$ with $y(0)=1$, you need the solution in two pieces. For $t < 2$, the equation becomes $y' + y = 0$, giving $y(t) = e^{-t}$ (using the initial condition). At $t=2$, the delta function creates a unit jump, so $y(2^+) = y(2^-) + 1 = e^{-2} + 1$. For $t > 2$, solving $y' + y = 0$ with this new initial condition gives $y(t) = (e^{-2} + 1)e^{-(t-2)} = e^{-t} + e^{2-t}$. Now, $\int_0^\infty y(t) \delta(t-3) dt$ uses the fundamental property of delta functions: $\int f(t)\delta(t-a) dt = f(a)$. Since we need $y(3)$ and $3 > 2$, we use the second piece: $y(3) = e^{-3} + e^{2-3} = e^{-3} + e^{-1}$. Answer A ($e^{-3}$) ignores the jump from the impulse. Answer C ($e^{-2} + e^{-1}$) incorrectly evaluates at $t=2$ instead of $t=3$. Answer D contains unnecessary unit step notation and doesn't properly account for the impulse effect. Key strategy: When solving impulse problems, always check which time interval your evaluation point falls in, and remember that delta function integrals simply evaluate the integrand at the impulse location.
17. A linear time-invariant system, initially at rest, has a response to a unit step input $u(t)$ given by $y_{step}(t) = (1 - e^{-2t})\cos(t)$. What is the system's response to a unit impulse input $\delta(t)$?
    Answer: $2e^{-2t}\cos(t) + (e^{-2t}-1)\sin(t)$ When you encounter a linear time-invariant (LTI) system problem relating step and impulse responses, remember that these responses are connected through differentiation: the impulse response is the derivative of the step response. For an LTI system, if $h(t)$ is the impulse response and $s(t)$ is the step response, then $h(t) = \frac{ds(t)}{dt}$. This fundamental relationship exists because the unit step is the integral of the unit impulse. Given $y_{step}(t) = (1 - e^{-2t})\cos(t)$, you need to find its derivative. Using the product rule: $h(t) = \frac{d}{dt}[(1 - e^{-2t})\cos(t)]$ First term: $\frac{d}{dt}(1 - e^{-2t}) = 2e^{-2t}$ Applying the product rule: $h(t) = 2e^{-2t}\cos(t) + (1 - e^{-2t})(-\sin(t))$ Simplifying: $h(t) = 2e^{-2t}\cos(t) + (e^{-2t} - 1)\sin(t)$ This matches answer D. Answer A is simply the step response itself—a common error of confusing step and impulse responses. Answer B represents only the first term of the correct derivative, missing the product rule's second term entirely. Answer C has the wrong sign on the sine term, likely from incorrectly applying the derivative of cosine. Study tip: Always remember that for LTI systems, impulse response = derivative of step response. When differentiating products involving exponentials and trigonometric functions, carefully apply the product rule and watch your signs—trigonometric derivatives are where sign errors commonly occur.
18. The solution to the initial value problem $y'' + 4y' + 5y = g(t)$ with $y(0)=0$ and $y'(0)=0$ has the Laplace transform $Y(s) = \frac{5e^{-2s}}{(s+2)^2+1}$. What is the forcing function $g(t)$?
    Answer: $5\delta(t-2)$ When you encounter a differential equation problem involving Laplace transforms, your goal is to work backward from the given transform to find the forcing function. The key insight is recognizing how the transform properties reveal information about the original function. Given $Y(s) = \frac{5e^{-2s}}{(s+2)^2+1}$, you need to identify what produces this specific form. The exponential factor $e^{-2s}$ is crucial—this represents the shifting property of Laplace transforms, which indicates a time delay or impulse at $t = 2$. The denominator $(s+2)^2+1$ would normally correspond to functions involving $e^{-2t}\sin(t)$ or $e^{-2t}\cos(t)$. However, the exponential $e^{-2s}$ in the numerator suggests we're dealing with a shifted impulse function rather than a continuous function. For an impulse function $\delta(t-a)$, the Laplace transform is $e^{-as}$. Here, $5\delta(t-2)$ has transform $5e^{-2s}$. When this impulse acts as the forcing function, the system's response involves the transfer function $\frac{1}{s^2+4s+5} = \frac{1}{(s+2)^2+1}$, giving exactly the form shown. Choice A) $\delta(t-2)$ is missing the factor of 5. Choice B) $5u(t-2)$ is a step function, not an impulse, and would produce $\frac{5e^{-2s}}{s}$ in the numerator. Choice C) represents a shifted exponential sine function, which would have a more complex transform structure. Remember: exponential factors $e^{-as}$ in Laplace transforms typically signal either time shifts in continuous functions or impulses at $t = a$. Check the overall structure to distinguish between these cases.
19. A system is modeled by $y''+9y=g(t)$ with $y(0)=0$ and $y'(0)=0$. The system response is observed to be $y(t) = \frac{4}{3}u(t-1)\sin(3(t-1))$ for $t \ge 0$. What is the forcing function $g(t)$?
    Answer: $4\delta(t-1)$ When you encounter a differential equation with a specific system response, you're working backwards from the solution to find the forcing function. This requires understanding how impulse functions and their responses relate. The key insight is recognizing what type of response $y(t) = \frac{4}{3}u(t-1)\sin(3(t-1))$ represents. This is the response to an impulse at $t=1$, not a continuous forcing function. The unit step function $u(t-1)$ indicates the response "turns on" at $t=1$, and the sine term $\sin(3(t-1))$ shows oscillation at the natural frequency $\omega = 3$ (from $y'' + 9y = 0$ where $\omega^2 = 9$). For the homogeneous equation $y'' + 9y = 0$, an impulse $\delta(t-1)$ produces a response that begins oscillating at $t=1$ with amplitude determined by the impulse strength. Since the impulse response has the form $\frac{A}{\omega}u(t-1)\sin(\omega(t-1))$ where $A$ is the impulse magnitude, we have $\frac{A}{3} = \frac{4}{3}$, giving us $A = 4$. Therefore, $g(t) = 4\delta(t-1)$, which is choice B. Choice A ($4u(t-1)$) would produce a step response, not an oscillatory response. Choice C ($\frac{4}{3}\sin(3(t-1))$) confuses the response with the forcing function. Choice D ($4\delta(t)$) places the impulse at the wrong time. Remember: when you see a suddenly-starting oscillatory response at the natural frequency, think impulse forcing at that moment, not continuous forcing.
20. The function $f(t) = \sum_{n=1}^{\infty} \frac{1}{n}\delta(t-n)$ represents impulses at positive integer values. What is the Laplace transform of the truncated version $g(t) = \sum_{n=1}^{3} \frac{1}{n}\delta(t-n)$?
    Answer: $\mathcal{L}\{g(t)\} = e^{-s} + \frac{e^{-2s}}{2} + \frac{e^{-3s}}{3}$ The correct answer is A. Using the linearity of the Laplace transform and the fact that $\mathcal{L}\{\delta(t-a)\} = e^{-as}$, we have $\mathcal{L}\{g(t)\} = \mathcal{L}\{\delta(t-1) + \frac{1}{2}\delta(t-2) + \frac{1}{3}\delta(t-3)\} = e^{-s} + \frac{1}{2}e^{-2s} + \frac{1}{3}e^{-3s}$. Choice B incorrectly factors out a common denominator. Choice C incorrectly includes a factor of $\frac{1}{s}$, which would arise from step functions, not delta functions. Choice D incorrectly squares the denominators.

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