# Convolution by differentiation property of convolution

This is the question when i am trying to do it by the deravative property of convolution https://en.wikipedia.org/wiki/Convolution#Differentiation like this Actual answer should come like this

$$y(t)= \begin{cases}0, & \text{for }t < 0 \\ 7-3t, & \text{for } 0<t<1 \\ 5-t, & \text{for } 1<t<5 \\ 7, & \text{for } t>5 \end{cases}$$

i am getting answer -6 at t=4 bu the actual answer should come 1

P.S:- Please dont give me alternative method to solve this i know it already instead tell me whats wrong i am doing in this method also i know i am violating the policy by putting the image but i will edit it lator in text form .

• Now you've got your answer :-) finally ! – Fat32 Aug 8 '17 at 14:02
• – msm Aug 10 '17 at 9:27
• $u(t)$ would need to be defined. – Olli Niemitalo Aug 28 '17 at 8:25
• $u(t)$ is standard unit step signal and $r(t)$ is unit ramp signal@OlliNiemitalo – Rohit Aug 30 '17 at 3:28

First of all let me state that, the property of differentiation under (convolution) integral operator may not be holding true for arbitrary signals but only for some subclass, such as absolutely integrable ones. That's stated in the wikipedia link OP provides about convolution. All I can make sense from that statement is that interchange of differentiation and integration is only possible if the integral is convergent. That's formal mathematics and I'm not very sure whether that applies to generalized functions, such as $u(t)$ and $\delta(t)$ as well. So I'll leave the discussion to mathematicians for a moment and proceed with the solution, treating the functions as if they're well behaving.

So far you've correctly obtained the signal $y(t)'$ from the differentiation of convolution property as: $$y(t)' = h(t)' \star x(t) = -3 u(t) + 2u(t-1) + u(t-5)$$

Now it comes to proper integration of $y(t)'$ to obtain $y(t)$ which was the most misleading step; $$\int_{-\infty}^{t} y(\tau)' d\tau = \int_{-\infty}^{t} -3 u(t) + 2u(t-1) + u(t-5) d\tau$$

the left side evaluates as;

$$\int_{-\infty}^{t} y(\tau)' d\tau = \int_{-\infty}^{t} dy d\tau = y(\tau)+C |_{-\infty}^{t} = (y(t)+C)-(y(-\infty)+C) = y(t)-y(-\infty)$$

Now here I shall make a few simple comments. As below we argue with @msm whether this is a definite integral or an indefinite one? He seems to claim it as an idefinite one with constant of integration at the output. (I don't actually know it because he doesn't put his equation there inspite of all my attempts, also he seems(?) to delete that comment as well? Rather he says it's simple to see. I don't know. It's always better to make it clear when requested to avoid any confusion. Of course this is a definite integral with clear limits on it, hence at the output you won't have a constant of integration whatsoever. The integrand is antiderivatived, and evaluated at the limits and subtracted so that the constant of integration is cancelled. What remains there as $y(-\infty)$ the integrand evaluated at the limits. For causal systems and signals applied, $y(-\infty)$ is by defintion zero, hence it's ignored, but when you have a noncausal system (like in the example system with $h(t) = u(-t-1)$ you may not ignore it and hence the term $y(-\infty)$ remains.

The right side of the equation is integrated as follows: $$= \int_{-\infty}^{t} -3 u(\tau) + 2u(\tau-1) + u(\tau-5) d\tau = \int_{0}^{t} -3 u(\tau) + 2u(\tau-1) + u(\tau-5) d\tau$$

$$= (-3 r(t) + 2r(t-1) + r(t-5) + C)-(-3 r(0) + 2r(0-1) + r(0-5) + C)$$ $$= -3 r(t) + 2r(t-1) + r(t-5)$$

Where we have indeed antiderivated each integrand and subtracted their evaluation at $\tau=t$ and $\tau=0$, the latter one all providing $0$

Hence we have; $$y(t) = -3 r(t) + 2r(t-1) + r(t-5) + y(-\infty)$$

At this point it's worth noting that, the output $y(t)$ depends on itself due to the $y(-\infty)$ term. Where you can fortunately (easily) evaluate it from the convolution integral as $$y(t) = \int_{-\infty}^{\infty} h(t-\tau)x(\tau)d\tau$$ as $t \rightarrow -\infty$ resulting in; $$y(t) = \int_{-\infty}^{\infty} h(-\infty-\tau)x(\tau)d\tau = y(t) = \int_{-\infty}^{\infty} 1 x(\tau)d\tau = 7$$

Hence the solution becomes, $$y(t) = 7 -3 r(t) + 2r(t-1) + r(t-5)$$ and finally $$y(4) = 7 - 3 r(4) + 2 r(3) + r(-1) = 7 - 3 \times 4 + 2 \times 3 + 1 \times 0 = 1$$

Below you can see the verification by direct evaluation of the colvolution integral.

(EDIT: see the first paragraph to get the point) As warned by the OP, that the differentiation property under the convolution is not valid for the signal $h(t) = u(-t-1)$, so we then resort to direct evaluation to check the answer.

$$y(t) = x(t) \star h(t) = x(t) \star u(-t-1)$$

The graphical method provides the easiest evaluation of this integral for particular output value of $t=4$ as:

$$y(4) = \int_{\tau= 5}^{\tau=6} x(\tau)h(4-\tau) d\tau = \int_{\tau= 5}^{\tau=6} 1 \cdot 1 d\tau = 1$$

So the requested value is: $y(4) = 1$ , hence you are right the answer is $y(4) = 1$ and (EDIT: see the first paragraph to get the point) [it seems the differentiation property is not applicable for arbitrary signals but those which are absolutely integrable.]

An example to see the diferentiation property is; $x(t) = u(t) - u(t-1)$ and $h(t) = u(t) - u(t-1)$ where we have $y(t)' = u(t) - 2u(t-1) + u(t-2)$ and the differentiation property yields: $$y(t)' = h(t)' \star x(t) = (\delta(t) - \delta(t-1)) \star x(t) = u(t) - 2u(t-2) + u(t-2)$$

Sorry for the confusion...

• sir i have one doubt wikipedia say this differention property applicable only when the both the signal should be absolute integrable but here unit step is not absolute integrable so how this property is applied here ? – Rohit Aug 7 '17 at 15:55
• Wait sorry sir the answer is not -6 its 1 because i getting 1 by general convolution method – Rohit Aug 7 '17 at 15:59
• sir the answer you are giving is not the correct answer, thats whats i am getting but thats wrong, answer should come 1 at t=4 i verified it by general convolution method – Rohit Aug 7 '17 at 16:05
• Down-voting for "the differentiation property under the convolution is not valid for the signal $h(t)=u(−t−1)$" which is wrong and the fact that the (originally wrong) answer does not address the OP's question that says "P.S:- Please dont give me alternative method to solve this i know it already instead tell me whats wrong i am doing in this method..." – msm Aug 8 '17 at 12:25
• @Fat32 sir first you got $y(t)-y(-\infty)$ then you should get the the value at $y(-\infty)$ but you have calcuted lator on $y(\infty)$ why i am really confused please help me? – Rohit Aug 9 '17 at 1:58