Given xc(t) = sin(40nt) + cos(80nt) a) Determine the CTFT of xe (t). b) x (t) is sampled with T, = 1/50 second. What is x[n]? c) Plot X (e) = DTFT (x[n])
Given xc(t) = sin(40nt) + cos(80nt) a) Determine the CTFT of xe (t). b) x (t) is sampled with T, = 1/50 second. What is x[n]? c) Plot X (e) = DTFT (x[n])
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![### Given
\[ x_c(t) = \sin(40\pi t) + \cos(80\pi t) \]
### Questions
#### a) Determine the CTFT of \( x_c(t) \).
#### b) \( x_c(t) \) is sampled with \( T_s = 1/50 \) second. What is \( x[n] \)?
#### c) Plot \( X(e^{j\omega}) = DTFT\{x[n]\} \).
---
### Solution:
#### a) Continuous-Time Fourier Transform (CTFT) of \( x_c(t) \)
To determine the CTFT of \( x_c(t) = \sin(40\pi t) + \cos(80\pi t) \), we use the known Fourier Transform pairs:
1. For \( \sin(40\pi t) \):
\[
\sin(40\pi t) \Rightarrow \frac{1}{2j} \left[ \delta(f - 20) - \delta(f + 20) \right]
\]
2. For \( \cos(80\pi t) \):
\[
\cos(80\pi t) \Rightarrow \frac{1}{2} \left[ \delta(f - 40) + \delta(f + 40) \right]
\]
Thus, the CTFT of \( x_c(t) \) is:
\[
X_c(f) = \frac{1}{2j} \left[ \delta(f - 20) - \delta(f + 20) \right] + \frac{1}{2} \left[ \delta(f - 40) + \delta(f + 40) \right]
\]
#### b) Sampling \( x_c(t) \) with \( T_s = 1/50 \) second
Using the sampling period \( T_s = 1/50 \) second, we can express the sampled signal \( x[n] \) as:
\[
x[n] = x_c(nT_s) = \sin(40\pi (n/50)) + \cos(80\pi (n/50))
\]
Simplify the arguments:
\[
x[n] = \sin\left(\frac{40\pi n}{50}\right) + \cos\left(\frac{80\pi n}{50](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76f42b50-a878-4748-80c4-72d452cd7a0f%2F0076a553-4303-4e33-b8ab-79c7da80213d%2Fdpvf2x9_processed.png&w=3840&q=75)
Transcribed Image Text:### Given
\[ x_c(t) = \sin(40\pi t) + \cos(80\pi t) \]
### Questions
#### a) Determine the CTFT of \( x_c(t) \).
#### b) \( x_c(t) \) is sampled with \( T_s = 1/50 \) second. What is \( x[n] \)?
#### c) Plot \( X(e^{j\omega}) = DTFT\{x[n]\} \).
---
### Solution:
#### a) Continuous-Time Fourier Transform (CTFT) of \( x_c(t) \)
To determine the CTFT of \( x_c(t) = \sin(40\pi t) + \cos(80\pi t) \), we use the known Fourier Transform pairs:
1. For \( \sin(40\pi t) \):
\[
\sin(40\pi t) \Rightarrow \frac{1}{2j} \left[ \delta(f - 20) - \delta(f + 20) \right]
\]
2. For \( \cos(80\pi t) \):
\[
\cos(80\pi t) \Rightarrow \frac{1}{2} \left[ \delta(f - 40) + \delta(f + 40) \right]
\]
Thus, the CTFT of \( x_c(t) \) is:
\[
X_c(f) = \frac{1}{2j} \left[ \delta(f - 20) - \delta(f + 20) \right] + \frac{1}{2} \left[ \delta(f - 40) + \delta(f + 40) \right]
\]
#### b) Sampling \( x_c(t) \) with \( T_s = 1/50 \) second
Using the sampling period \( T_s = 1/50 \) second, we can express the sampled signal \( x[n] \) as:
\[
x[n] = x_c(nT_s) = \sin(40\pi (n/50)) + \cos(80\pi (n/50))
\]
Simplify the arguments:
\[
x[n] = \sin\left(\frac{40\pi n}{50}\right) + \cos\left(\frac{80\pi n}{50
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