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I want to find the impulse response of the linear time invariant system given by

$y''(t)+4y'(t)+3y(t)=x'(t)+2x(t)$

using time domain analysis.

As solved in the textbook "Continuous and Discrete Signals and Systems" by Soliman and Srinath (1990, Prentice Hall International), Section 2.5, page 86, the impulse response is assumed

$h(t)=(C_1e^{-3t}+C_2e^{-t})u(t)$

Then this is plugged into the equation $h''(t)+4h'(t)+3h(t)=\delta'(t)+2\delta(t)$ to yield $C_1=0,C_2=1.$

However, when I apply the Simplified Impulse Matching Method as prescribed by the text "Linear Systems and Signals" by Lathi and Green (2017, Oxford University Press), section 2.3, page 165, the zero input solution of the differential equation is

$y_n(t)=c_1e^{-3t}+c_2e^{-t}$

and the constants are found from the initial conditions

$y_n(0)=0,y'_n(0)=1$

which gives

$y_n(t)=-\dfrac{1}{2}e^{-3t}+\dfrac{1}{2}e^{-t}$

Then the impulse response is found by plugging $y_n(t)$ into the right hand side of the diff. eq. and multiplying by $u(t)$, that is

$(y'_n(t)+2y_n(t))u(t)$

which gives

$\left(\dfrac{1}{2}e^{-3t}+\dfrac{1}{2}e^{-t}\right)u(t)$

as the impulse response. What am I missing that is causing the two answers to not match? Thanks in advance for any help!

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1 Answer 1

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I don't see how the first method would result in $c_1=0$ and $c_2=1$. With

$$h(t)=\big[c_1e^{-3t}+c_2e^{-t}\big]u(t)$$

you obtain

$$h'(t)=\big[\ldots\big]u(t)+(c_1+c_2)\delta(t)$$

and

$$h''(t)=\big[\ldots\big]u(t)+(-3c_1-c_2)\delta(t)+(c_1+c_2)\delta'(t)$$

I left out the terms multiplied with $u(t)$ because they don't matter for the determination of the constants $c_1$ and $c_2$.

Plugging this into the differential equation gives

$$(c_1+c_2)\delta'(t)+(c_1+3c_2)\delta(t)=\delta'(t)+2\delta(t)$$

resulting in

$$\begin{align}&c_1+c_2=1\\&c_1+3c_2=2\end{align}$$

which corresponds to the solution of the second method.

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  • $\begingroup$ Thanks for your response! I see, I did not simplify $(c_1e^{-3t}+c_2e^{-t})\delta(t)$ to $(c_1+c_2)\delta(t)$ in the expression for $h'(t)$ and got an extra $(-3c_1-c_2)\delta(t)$ term because of that! $\endgroup$ Mar 7, 2023 at 18:18
  • $\begingroup$ Could you possibly shed some light on why the book goes on to do the same as me? Is there any delicate point or is it just a mistake? $\endgroup$ Mar 7, 2023 at 18:21
  • $\begingroup$ @kobaltishblue_312: No idea about the book, but if their solution is $c_1=0$ and $c_2=1$ then it's simply wrong. $\endgroup$
    – Matt L.
    Mar 7, 2023 at 18:37

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