MHB Compute Discrete Time Fourier Transform

AI Thread Summary
The discussion focuses on computing the Discrete Time Fourier Transform (DTFT) of the signal x_n = a^n cos(λ₀n)u_n, where |a| < 1 and u_n is the unit step function. The user successfully applies Euler's formula to express the cosine term and sets up the DTFT summation, but encounters difficulties in simplifying the expression further. Another participant suggests re-evaluating the substitution and highlights the modulation property, indicating that the DTFT can be represented in terms of shifted frequency components. The conversation emphasizes the importance of correctly applying these properties to proceed with the computation.
nacho-man
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Hi bros,

so I feel like I am very close, but cannot find out how to go further.

Q.1 Compute the DTFT of the following signals, either directly or using its properties (below a is a fixed constant |a| < 1):

for $x_n = a^n \cos(\lambda_0 n)u_n$ where $\lambda_0 \in (0, \pi)$ and
$u_n$ is the step function, i.e $u_n = 1 $ for $n\ge 0 $ and $0$ otherwise.so,

$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} x_n e^{-i\lambda n}$$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} a^n \cos(\lambda_0 n) e^{-i\lambda n}$

using euler's formula: $\cos(\lambda_0 n)$ = $\frac{e^{i\lambda_0 n}+ e^{-i\lambda_0 n}}{2} $

so $X(e^{i \lambda}) = \frac{1}{2} \sum_{n=0}^{+\infty} (a e^{-i \lambda}( e^{i\lambda_0}+ e^{-i\lambda_0}))^n $

which gives

$\frac{1}{2} \sum_{n=0}^{+\infty} a^n(e^{i \lambda_0 - i\lambda} + e^{-i \lambda_0 - i\lambda})^n)$

and now i am stuck
.
i think i have it right upto this point, but i do not know how to proceed.

also in our notes, he has said to use the property called modulation, which meanswhere we have $x^n e^{i \lambda_0 n}$ the DTFT will be of the form $X(e^{i(\lambda-\lambda_0)})$ANY HELP IS APPRECIATED! thank you!
 
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Hi nacho! :)

nacho said:
$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} a^n \cos(\lambda_0 n) e^{-i\lambda n}$

using euler's formula: $\cos(\lambda_0 n)$ = $\frac{e^{i\lambda_0 n}+ e^{-i\lambda_0 n}}{2} $

so $X(e^{i \lambda}) = \frac{1}{2} \sum_{n=0}^{+\infty} (a e^{-i \lambda}( e^{i\lambda_0}+ e^{-i\lambda_0}))^n $

Not so fast.
Let's first do only the substitution.

$$X(e^{i \lambda}) = \sum_{n=0}^{+\infty} a^n \cdot \frac{1}{2} (e^{i\lambda_0 n}+ e^{-i\lambda_0 n}) \cdot e^{-i \lambda n} $$

This is different from what you have. :eek:
which gives

$\frac{1}{2} \sum_{n=0}^{+\infty} a^n(e^{i \lambda_0 - i\lambda} + e^{-i \lambda_0 - i\lambda})^n)$

and now i am stuck
.
i think i have it right upto this point, but i do not know how to proceed.

Let's redo that and simplify to:
$$X(e^{i \lambda})
= \frac{1}{2} \sum_{n=0}^{+\infty} a^n (e^{-i(\lambda - \lambda_0) n}+ e^{-i(\lambda + \lambda_0) n})
= \frac{1}{2} \sum_{n=0}^{+\infty} a^n e^{-i(\lambda - \lambda_0) n} + \frac{1}{2} \sum_{n=0}^{+\infty} a^n e^{-i(\lambda + \lambda_0) n}
$$
also in our notes, he has said to use the property called modulation, which meanswhere we have $x^n e^{i \lambda_0 n}$ the DTFT will be of the form $X(e^{i(\lambda-\lambda_0)})$

Can you apply this now? (Wondering)
 
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