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How do we define convolution of: $$x(t)= \mathrm{sinc}(W t)$$ and $$y(t)= -j\mathrm{sinc}(2W t)$$

Answer: In the frequency domain, both of them are rectangular functions and multiplication of them is equal to the rectangular signal with smaller bandwidth i.e $x(t)$, which means convoluted signal is $\mathrm{sinc}(W t)$ with some scaling factor.

But I do not understand how this could be done, the two signals are orthogonal (because one is real and the other imaginary). So shouldn't convolution be 0? I would appreciate if someone could explain.

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You are confusing several different notions. First of all, two signals $x(t)$ and $y(t)$ are orthogonal if

$$\int_{-\infty}^{\infty}x(t)y^*(t)dt=0\tag{1}$$

From (1) it is clear that if two real-valued functions $x(t)$ and $y(t)$ are not orthogonal, i.e. the integral in (1) is not equal to zero, then also $x(t)$ and $jy(t)$ are not orthogonal. So your argument that the two given signals must be orthogonal because one of them is real-valued and the other one is imaginary is wrong.

Furthermore, even if two signals are orthogonal to each other, their convolution is usually non-zero. Just take two rectangular functions that do not overlap. Obviously, they are orthogonal, but their convolution is a triangular function.

Consequently, one way to interpret the convolution of the two sinc functions is that a low pass signal with cut-off frequency $W$ is filtered by a low pass filter with cut-off frequency $2W$, and multiplied by the (irrelevant) factor $-j$. Since the low pass filter has no effect on the low pass signal, the only change to the signal is the constant factor $-j$.

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  • $\begingroup$ Thank you for clearing my concepts as well as for answering the question. $\endgroup$ – Curious91 Jan 9 '15 at 20:19
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You can treat the $-j$ in $y(t)$ as a simple scaling factor. Let's redefine $y(t)=\mathrm{sinc}(2Wt)$. Now we want $$ x(t)\otimes -jy(t)= \int_{-\infty}^\infty -jx(\tau)y(t-\tau)d\tau=-j\int_{-\infty}^\infty x(\tau)y(t-\tau)d\tau= -j\big[x(t)\otimes y(t)\big]$$

As you have already noted the Fourier transform of the sinc() is a rectangular pulse, so the result of multiplication in the frequency domain is just a rectangular pulse with the smaller bandwidth - note you do have to take into account the amplitude scaling since: $$ \mathrm{sinc}(2Wt) \rightleftharpoons \frac{1}{2W}\mathrm{rect}\bigg(\frac{f}{2W}\bigg) $$

Given this information and what Matt provided in his answer, you should be able to easily find your answer.

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