So I attempted drawing the new signal below. It is reflected, shifted left pi/3 ... but I don\'t know how the exponential signal (alpha^n) is supposed to affect the signal back to the frequency domain!!! Please be clear in solution. =)
Sketch the DTFT of the sequence y(n) = x(- n) e-jn x /3 + alpha n for all n (a is a constant) Thise is the sketch without the constant alpha n :
Solution
due to the 2?-periodicity of the DTFT, the integral in figure given can be computed over any 2?-wide interval on the real line (i.e. between 0 and 2?, for instance). The relation between a sequence x[n] and its DTFT X(exp(j?)) will be indicated in the general case by
x [n] == X (e^j? )
Clearly, for the DTFT to exist, the sum must converge.
.
So I attempted drawing the new signal below- It is reflected- shifted.docx
1. So I attempted drawing the new signal below. It is reflected, shifted left pi/3 ... but I don't know
how the exponential signal (alpha^n) is supposed to affect the signal back to the frequency
domain!!! Please be clear in solution. =)
Sketch the DTFT of the sequence y(n) = x(- n) e-jn x /3 + alpha n for all n (a is a constant) Thise
is the sketch without the constant alpha n :
Solution
due to the 2?-periodicity of the DTFT, the integral in figure given can be computed over any 2?-
wide interval on the real line (i.e. between 0 and 2?, for instance). The relation between a
sequence x[n] and its DTFT X(exp(j?)) will be indicated in the general case by
x [n] == X (e^j? )
Clearly, for the DTFT to exist, the sum must converge.
![So I attempted drawing the new signal below. It is reflected, shifted left pi/3 ... but I don't know
how the exponential signal (alpha^n) is supposed to affect the signal back to the frequency
domain!!! Please be clear in solution. =)
Sketch the DTFT of the sequence y(n) = x(- n) e-jn x /3 + alpha n for all n (a is a constant) Thise
is the sketch without the constant alpha n :
Solution
due to the 2?-periodicity of the DTFT, the integral in figure given can be computed over any 2?-
wide interval on the real line (i.e. between 0 and 2?, for instance). The relation between a
sequence x[n] and its DTFT X(exp(j?)) will be indicated in the general case by
x [n] == X (e^j? )
Clearly, for the DTFT to exist, the sum must converge.](https://image.slidesharecdn.com/soiattempteddrawingthenewsignalbelow-itisreflected-shifted-230126013620-9911fba0/85/So-I-attempted-drawing-the-new-signal-below-It-is-reflected-shifted-docx-1-320.jpg)
