Math Problem Statement

solve the folloing question with accuracy in answer

For the given frequency spectrum of X(f) and Y(f), (a) Find Y(t) as a function of X(t). (b) If Z(t) = Y(t)Cos(2π * 30000t), sketch the frequency spectrum of Z(t).

Solution

Let's solve this problem step by step.

Question (a): Find Y(t)Y(t) as a function of X(t)X(t)

The relationship between X(f)X(f) and Y(f)Y(f) suggests a frequency shift because the frequency spectrum of Y(f)Y(f) has shifted components compared to X(f)X(f).

  • Given Observations:

    1. X(f)X(f) is a triangular pulse spanning from 3400-3400 Hz to 34003400 Hz, with a peak magnitude of 10 at 0 Hz.
    2. Y(f)Y(f) has its components shifted to center around ±10kHz\pm 10 \, \text{kHz} and reduced in amplitude. The peak magnitude is 12, and the triangle width is smaller.
  • Frequency Shift: Y(f)Y(f) is derived from X(f)X(f) by shifting X(f)X(f) to ±10kHz\pm 10 \, \text{kHz} and scaling. This implies: Y(f)=12X(f±10,000)Y(f) = 12 \cdot X(f \pm 10,000)

  • Time Domain Representation: A shift in the frequency domain corresponds to a multiplication in the time domain by a complex exponential: Y(t)=12X(t)cos(2π10,000t)Y(t) = 12 \cdot X(t) \cdot \cos(2\pi \cdot 10,000t) This result is the modulated version of X(t)X(t) by a carrier frequency at 10kHz10 \, \text{kHz}.

Question (b): Sketch the frequency spectrum of Z(t)Z(t), where

Z(t)=Y(t)cos(2π30,000t)Z(t) = Y(t) \cos(2\pi \cdot 30,000t)

Step 1: Understand Z(t)Z(t)'s Modulation

Modulating Y(t)Y(t) by cos(2π30,000t)\cos(2\pi \cdot 30,000t) results in another frequency shift. In the frequency domain, the modulation by cos(2π30,000t)\cos(2\pi \cdot 30,000t) corresponds to splitting Y(f)Y(f) into two shifted components:

  • Y(f)Y(f) is shifted by +30,000Hz+30,000 \, \text{Hz} and 30,000Hz-30,000 \, \text{Hz}.

Step 2: Analyze the Spectrum

The original Y(f)Y(f) consists of two triangular pulses centered at ±10kHz\pm 10 \, \text{kHz}. Modulating Y(f)Y(f) by cos(2π30,000t)\cos(2\pi \cdot 30,000t) produces:

  1. One set of triangular pulses centered at ±40kHz\pm 40 \, \text{kHz} (from +10,000+30,000+10,000 + 30,000).
  2. Another set of triangular pulses centered at ±20kHz\pm 20 \, \text{kHz} (from 10,000+30,000-10,000 + 30,000).

Step 3: Final Spectrum

The new frequency spectrum Z(f)Z(f) will have:

  • Two triangular components centered at ±40kHz\pm 40 \, \text{kHz}, each scaled by the factor from Y(f)Y(f).
  • Two triangular components centered at ±20kHz\pm 20 \, \text{kHz}.

Final Answers:

(a) Y(t)=12X(t)cos(2π10,000t)Y(t) = 12 \cdot X(t) \cdot \cos(2\pi \cdot 10,000t)

(b) The frequency spectrum Z(f)Z(f) consists of:

  • Two triangular pulses centered at ±40kHz\pm 40 \, \text{kHz}.
  • Two triangular pulses centered at ±20kHz\pm 20 \, \text{kHz}.

Let me know if you'd like detailed sketches or have further questions! Here's how we can expand:

  1. Why does a frequency shift in X(f)X(f) cause modulation in X(t)X(t)?
  2. How can amplitude scaling in X(f)X(f) affect X(t)X(t)?
  3. What are the advantages of frequency modulation in communication systems?
  4. What is the relationship between bandwidth and the width of the triangular pulses in this problem?
  5. How do you mathematically verify the result for Z(t)Z(t)?

Tip: Always ensure the units are consistent (Hz vs kHz) while analyzing spectra.

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Math Problem Analysis

Mathematical Concepts

Fourier Transform
Frequency Modulation
Frequency Shifting
Time and Frequency Domain

Formulas

Y(t) = X(t) * cos(2π * fc * t) (frequency modulation)
Frequency shift in time domain ↔ multiplication by complex exponential in frequency domain

Theorems

Fourier Transform Duality
Frequency Shifting Theorem

Suitable Grade Level

Undergraduate (Electrical/Communication Engineering)