Why exponential is ignored in particular solution for impulse response ??

For a system govern by the equation: $$2y'(t) +4y(t) =3x(t)$$

To calculate it's impulse response we replace $y(t)$ with $h(t)$ and $x(t)$ with $\delta(t)$ and get

$2h'(t)+4h(t)=3\delta(t)$ which's homogeneous part of the solution is $h(t)=C e^{-2t}u(t)$

To find C we get that : $2\frac{d}{dt}(Ce^{-2t}u(t))+4Ce^{-2t}u(t) = 3\delta(t)$

$Ce^{-2t}\delta(t)=\frac{3}{2}\delta(t)$

$C=\frac{3}{2}e^{2t}$

Which should be wrong cause value of a constant should not have any value containing $t$ . Where did I do wrong ?

In book last two line is not provided but when the use the value only use $1.5$ instead of $1.5e^{2t}$ . Is that any particular point did I miss? ? Or do something basic thing wrong ?

• I have given a schematic outline of what you should do. Say if you still don't understand and would like to see more of the derivation. – Chappers Jun 18 '15 at 11:24

If the impulse is at time $t_0$, you need to consider the delta function $\delta(t-t_0)$. You then evaluate the limit of the derivative either side of $t=t_0$ to find the constant, which you should expect to depend on $t_0$ (or how would the derivative be able to jump specifically at $t_0$?).

Suppose you have an impulse at $t=\tau$. Then you are solving the equation $$2h'+4h = \delta(t-\tau). \tag{1}$$ Since the independent variable is time, we are normally interested in the boundary conditions $h(t)=0$ for $t<\tau$ (it is clear that this is a solution of (1), since the right-hand side is zero for $t \neq \tau$.

For $t>\tau$, we need to have a solution to $$2h'+4h = 0, \tag{2}$$ but it needs to be consistent with the "jump condition" at $t=\tau$, that we find by integrating (1) over a small interval including $\tau$: if you think about it, the nature of a discontinuity of a function will get worse as we take derivatives; on the other hand if $h$ has a finite discontinuity, its integral will be continuous. Hence, we have $$\int_{\tau-\varepsilon}^{\tau+\varepsilon} (2h'(t)+4h(t)) \, dt = \int_{\tau-\varepsilon}^{\tau+\varepsilon} \delta(t-\tau) \, dt = 1,$$ and the left-hand side, by the above discussion, should be $$\int_{\tau-\varepsilon}^{\tau+\varepsilon} 2h'(t) \, dt = 2(h(\tau+\varepsilon)-h(\tau-\varepsilon)).$$ Taking $\varepsilon \to 0$, we normally write this as $$\left[2h(\tau)\right]_{-}^{+} = 1$$

In this case, $h(\tau^-)=0$, so we need to have $2h(\tau^+)=1$. The general solution to (2) is $Ce^{2t}$, so the jump condition implies that we need $$Ce^{2\tau} = \frac{1}{2},$$ or $C=\frac{1}{2}e^{-2\tau}$, so the solution to (1) is $$h(t) = \begin{cases} 0 & t<0 \\ \frac{1}{2}e^{2(t-\tau)} & t>0 \end{cases}$$

Now, if you really do want to just consider an impulse at $t=0$, put $\tau=0$ into the above.

• I think $t_o =0$ as it's only mentioned that $y(t)=0 , -\infty <t<0$ and replace $y(t)$ with $h(t)$ and $x(t)$ with $\delta(t)$ – Anklon Jun 18 '15 at 11:28
• I have added an explanation of the general result above. – Chappers Jun 18 '15 at 13:45

$$Ce^{-2t}\delta(t)=\frac{3}{2}\delta(t) \\ C=\frac{3}{2}e^{2t}$$ [...] Or do something basic thing wrong ?

Yes, you did. When can you deduce from $ab=ac$ that $b=c$?

• Do you indicating that as for $t>0 , \delta(t)=0$ . So I should only consider the value for $t=0$ ? – Anklon Jun 18 '15 at 11:37

The left hand equation: The property of delta function: $$C\exp(-2t)\delta(t)=C\exp(-2\cdot0)\delta(t)=C\delta(t)$$

By joining above equation to the right hand equation: $$C\delta(t)=3/2\delta(t)$$ $$C = 3/2$$