Frequency Response Homework: Impulse & Cosine Output

Click For Summary
SUMMARY

The discussion focuses on solving a homework problem related to a causal FIR filter defined by the coefficients {bk} = {1, 4, 5, 4, 1}. The impulse response is determined as h[n] = δ[n] + 4δ[n-1] + 5δ[n-2] + 4δ[n-3] + δ[n-4]. The frequency response is derived using Euler’s formula, resulting in H(e^{jω}) = e^{-j2ω} [2cos(2ω) + 8cos(ω) + 5]. The final part of the problem involves substituting ω = 0.5π to find the output y[n] when the input is x[n] = cos(0.5πn).

PREREQUISITES
  • Understanding of FIR filters and their properties
  • Knowledge of impulse response and frequency response concepts
  • Familiarity with Euler's formula in signal processing
  • Basic skills in discrete-time signal analysis
NEXT STEPS
  • Study the derivation of impulse responses for various FIR filter coefficients
  • Learn about the application of Euler's formula in frequency response analysis
  • Explore the relationship between frequency response and output signals in discrete systems
  • Investigate the Nyquist rate and its implications in signal processing
USEFUL FOR

Students and professionals in electrical engineering, particularly those focusing on signal processing, filter design, and discrete-time systems analysis.

freezer
Messages
75
Reaction score
0

Homework Statement



3. (12 pts) Consider the causal FIR filter with {bk } = {1, 4, 5, 4, 1}.
(a) What is the impulse response of this filter (in terms of delta functions)?
(b) What is the frequency response of this filter? Simplify using Euler’s inverse formula.
(c) What is the output y[n] of this system when the input is x[n] = cos(0.5πn)?

Homework Equations


The Attempt at a Solution



(a) h[n]= δ[n]+ 4δ[n-1]+ 5δ[n-2]+ 4δ[n-3]+δ[n-4]

(b)

<br /> H(e^{j\hat{\omega}}) = 1 + 4e^{-j\hat{\omega}}+5e^{-j2\hat{\omega}}+4e^{-j3\hat{\omega}}+e^{-j4\hat{\omega}}<br />

<br /> H(e^{j\hat{\omega}}) = e^{-j2\hat{\omega}} [e^{j2\hat{\omega}} + e^{-j2\hat{\omega}}+4e^{j\hat{\omega}}+e^{-j\hat{\omega}}+5]<br />

<br /> H(e^{j\hat{\omega}}) = e^{-j2\hat{\omega}} [2cos(2\hat{\omega})+8cos(\hat{\omega})+5]<br />

(c) Does 0.5pi get substituted for omega hat to solve part c?
 
Physics news on Phys.org
Hopefully someone can give me a hint on this... It seems to me that I need an fs to complete this problem. omega hat = omega*Ts. If you are not given fs, do you just use the Nyquist rate?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
31
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K