If $x(t)=\sin(\frac{\pi t}{4})$ and $y(t)=\cos(\frac{\pi t}{4})$ then i need to find the Convolution $$z(t)=x(t) \circledast y(t)$$

So convolution will be

$$\begin{align} z(t) &=\int_{-\infty}^{\infty}x(\tau)y(t-\tau)d\tau \\ \\ &=\int_{-\infty}^{\infty}\sin\left(\frac{\pi \tau}{4}\right)\cos\left(\frac{\pi (t-\tau)}{4}\right)d\tau \\ \\ &= \tfrac12 \cos\left(\frac{\pi t}{4}\right)\int_{-\infty}^{\infty}\sin\left(\frac{\pi \tau}{2}\right)d\tau+\sin\left(\frac{\pi \tau}{4}\right)\int_{-\infty}^{\infty}\sin^2\left(\frac{\pi t}{4}\right)d\tau \\ \end{align}$$

now first integral will come $0$ because odd signal and 2nd integral value will come $\infty$ that is undefined.That means convolution is not possible for these signals.

But if i do it by fourier transform then i will get $$\begin{align} Z(j\omega) &= X(j\omega)Y(j\omega) \\ &= \frac{\pi^2}{j}( \delta(\omega- \tfrac{\pi}{4})-\delta(\omega+\tfrac{\pi}{4}) ) \\ &= 2\pi^2\sin \left(\frac{\pi t}{4} \right) \\ \end{align}$$

How its possible that i can get it by fourier transform but not by direct convolution?

However the answer given is $4\sin(\frac{\pi t}{4})$.

What is the mistake i am doing please help me?

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    $\begingroup$ i think that the answer given is wrong. you need to consider the difference between a finite energy signal and a finite power signal. and there are different definitions of the inner product between finite energy signals and for finite power signals. but i have never seen convolution to use the finite power definition of the inner product. $\endgroup$ – robert bristow-johnson Sep 30 '17 at 5:52
  • $\begingroup$ @robertbristow-johnson sir can i do it by fourier transform? $\endgroup$ – Rohit Sep 30 '17 at 7:29
  • $\begingroup$ you have a problem doing this with Fourier Transform because you are multiplying two dirac impulses against each other. can you tell us what the product of $\delta(t) \times \delta(t)$ is? $\endgroup$ – robert bristow-johnson Sep 30 '17 at 18:15
  • $\begingroup$ @robertbristow-johnson Sir it will be $\delta(0)\times\delta(t)$, i made that mistakes,multiplied directly.....Thanks $\endgroup$ – Rohit Oct 1 '17 at 1:21
  • $\begingroup$ what is $\delta(0)$? $\endgroup$ – robert bristow-johnson Oct 1 '17 at 1:25

This is a theoretical question without much practical interest but still it can be nice to check the results and investigate if intuition still holds.

First of all, if the convolution is regarded an LTI operation between an input $x(t)=\cos(\omega_0 t)$ and a system $y(t)=h(t)=\sin(\omega_o t)$ then it's immediately obvious that since $h(t)$ is an unstable system, the output can be unbounded even for a bounded input signal.

Furthermore assuming that Fourier transforms of the input, the system and the output exists (the integrals converge!) then we can assert the property that: $$ x(t) \star y(t) \longleftrightarrow X(\omega)Y(\omega) $$

When the sinusodal signals $x(t)=\cos(\omega_0 t)$ and $y(t)=\sin(\omega_0 t)$ are considered, it's obvious that their Fourier integrals do not converge. Hence their formal Fourier transforms do not exist. The solution is the acceptance of the generalised impulse function to represent the Fourier transform of the sinusodial signals; i.e., $$ x(t)=\cos(\omega_0 t) \longleftrightarrow X(\omega) = \pi [\delta(\omega-\omega_0) + \delta(\omega+\omega_0)]$$ and $$ y(t)=\sin(\omega_0 t) \longleftrightarrow Y(\omega) = \frac{\pi}{j} [\delta(\omega-\omega_0) - \delta(\omega+\omega_0)]$$

Then we consider that the above theorem still holds for the impulse functions as well and apply it here: $$ z(t)=\sin(\omega_0 t) \star \cos(\omega_0 t) \longleftrightarrow Z(\omega)=\frac{\pi}{j}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)] \cdot \pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)] $$

Using the impulse sifting property; $$f(x)\delta(x-a) = f(a)\delta(x-a)$$ we shall perform th multiplications as follows: $$ Z(\omega)= \frac{\pi^2}{j}[ \delta(\omega-\omega_0) \delta(\omega-\omega_0) + \delta(\omega-\omega_0) \delta(\omega+\omega_0) - \delta(\omega+\omega_0) \delta(\omega-\omega_0) - \delta(\omega+\omega_0) \delta(\omega+\omega_0)] $$

$$ Z(\omega)= \frac{\pi^2}{j}[ \delta(0) \delta(\omega-\omega_0) + \delta(-2\omega_0) \delta(\omega+\omega_0) - \delta(2\omega_0) \delta(\omega-\omega_0) - \delta(0) \delta(\omega-\omega_0)] $$

Now noting that $\delta(2\omega_0)=\delta(-2\omega_0)= 0$ those terms cancel and $\delta(0)=\infty$ terms remain, hence

$$ Z(\omega)= \frac{\pi^2}{j}[ \delta(0) \delta(\omega-\omega_0) - \delta(0) \delta(\omega-\omega_0)] $$

$$ Z(\omega)= \pi \delta(0) \frac{\pi}{j} [ \delta(\omega-\omega_0) - \delta(\omega-\omega_0)] $$

Which is recognized as the Fourier transform of the infinite amplitude sine wave as: $$ \boxed{ z(t) = \pi \delta(0) \sin(\omega_0 t) } $$

The conclusion is that the convolution between $x(t)=\sin(\omega_0 t)$ and $y(t)=\sin(\omega_0 t)$ produces an infinite amplitude sinudoidal wave of the same frequency $\omega_0$.

The time domain verification is as follows:

$$x(t) \star y(t) = \int_{-\infty}^{\infty} x(t-\tau)y(\tau) d\tau \leftrightarrow z(t) = \int_{-\infty}^{\infty} \cos(\omega_0(t-\tau)) \sin(\omega_0 \tau) d\tau$$

Using the trigonometric identity of $$\cos(x)\sin(y) = 0.5[\sin(y+x) + \sin(y-x)]$$ we can break the integral into two, where $x = \omega_0(t-\tau)$ and $y=\omega_0\tau$ , hence $$z(t) = 0.5 \int_{-\infty}^{\infty} \sin(\omega_0(t-\tau)+\omega_0 \tau) + \sin(\omega_0 \tau - \omega_0(t-\tau) ) d\tau$$

$$z(t) = 0.5 \int_{-\infty}^{\infty} \sin(\omega_0 t)d\tau + 0.5 \int_{-\infty}^{\infty} \sin(2\omega_0 \tau - \omega_0 t) d\tau$$

Now the first integral becomes $$ 0.5 \sin(\omega_0 t) \int_{-\infty}^{\infty} d\tau$$ while the second integral can be shown to be zero after a suitable change of variables, assuming for a given (fixed) t, $\phi = 2\omega_0 \tau - \omega_0 t$ and $d\phi = 2\omega_0 d\tau$ then the second integral becomes $ \frac{1}{2\omega_0} \int_{-\infty}^{\infty} \sin(\phi) d\phi = 0$.

Theferore the result of the convolution is: $$z(t) = \left( 0.5 \int_{-\infty}^{\infty} 1 d\tau \right)\sin(\omega_0 t) $$ Note that the integral that weights the sine wave has infinte value, moreover it can be shown that by the forward and inverse Fourier transform pairs one can recognize the integral as the forward Fouier transform $H(\omega)$ (evaluated at $w=0$, $H(0)$) of the signal $x(t)=1$ which is:

$$\mathcal{F} \{ 1\} = \int_{-\infty}^{\infty} 1 e^{-j \omega t} dt \equiv 2\pi \delta(\omega) $$ and therefore setting $w=0$ yields $$\int_{-\infty}^{\infty} 1 dt \equiv 2\pi \delta(0) $$

Finally plugging this into the result yields; $$\boxed{ z(t) = \pi \delta(0) \sin(\omega_0 t) }$$ which is the same as the result obtained from the Fourier method.

  • 1
    $\begingroup$ Thank you so much sir ...i was waiting for your answer :-) $\endgroup$ – Rohit Sep 30 '17 at 15:36
  • $\begingroup$ OK @Rohit ! But I start feeling as if your private tutor here ! $\endgroup$ – Fat32 Sep 30 '17 at 17:53
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    $\begingroup$ I am sorry sir... i didn't say that in any wrong sense..! $\endgroup$ – Rohit Sep 30 '17 at 17:57
  • $\begingroup$ I just asked your help because you can explain it in better way $\endgroup$ – Rohit Sep 30 '17 at 17:58
  • 1
    $\begingroup$ ok Rohit no problem :-)) $\endgroup$ – Fat32 Sep 30 '17 at 18:08

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