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Chapter 5: Problem 2

The inverse of the period \(\mathrm{T}\) is called the "temporal frequency" of the cosinusoidal signal and is given the symbol \(f\); the units of \(f=\frac{1}{T}\) are (seconds) \(^{-1}\) or hertz \((\mathrm{Hz})\). Write \(x(t)\) in terms of \(f .\) How is \(f\) related to \(\omega\) ? Explain why \(f\) gives the number of cycles of \(x(t)\) per second.

Short Answer

Expert verified
The signal is \( x(t) = A \cos(2\pi f t + \phi) \). The temporal frequency \( f = \frac{\omega}{2\pi} \) and it represents the number of cycles per second.

Step by step solution

01

Identify the given relationship

The inverse of the period (T) is called the temporal frequency (f), given by the formula: \[ f = \frac{1}{T} \]
02

Express the cosine signal with period

The standard form of the cosinusoidal signal is: \[ x(t) = A \cos(\omega t + \phi) \] where \(\omega\) is the angular frequency (in radians per second).
03

Relate period T to angular frequency \(\omega\)

The period (T) is related to the angular frequency by: \[ T = \frac{2\pi}{\omega} \]
04

Substitute the period into the temporal frequency

Using the relation \( T = \frac{2\pi}{\omega} \), substitute into the formula for temporal frequency: \[ f = \frac{1}{T} = \frac{1}{\frac{2\pi}{\omega}} = \frac{\omega}{2\pi} \]
05

Rewrite the cosinusoidal signal in terms of \(f\)

Since \( \omega = 2\pi f \), substitute into the signal: \[ x(t) = A \cos(2\pi f t + \phi) \]
06

Explain temporal frequency in terms of cycles per second

The temporal frequency (f) denotes the number of cycles the cosinusoidal signal completes per second. Since one complete cycle corresponds to a period T, and \( f = \frac{1}{T} \), this means in one second, the signal completes \( f \) cycles.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cosinusoidal Signal
A cosinusoidal signal is a type of periodic signal that can be expressed using the cosine function. The general form of a cosinusoidal signal in time domain is: \[ x(t) = A \, \cos(\omega t + \phi) \]
Here,
  • \(A\) is the amplitude, representing the peak value of the signal.
  • \(\omega\) is the angular frequency in radians per second, depicting how fast the signal oscillates.
  • \(t\) is the time variable.
  • \(\phi\) is the phase, determining the signal's initial angle or position at time \(t = 0\).
This formula succinctly captures the essence of a cosinusoidal signal, making it an essential concept in signal processing and communications.
Angular Frequency
Angular frequency, represented by \(\omega\), is crucial in understanding how rapidly a cosinusoidal signal cycles. Angular frequency is linked to temporal frequency \(f\) through the relation: \[ \omega = 2\pi f \]
This means that the angular frequency is essentially the temporal frequency scaled by \(2\pi\).
  • Angular frequency \(\omega\) provides a measure of rotation speed in radians per second.
  • It's derived from the temporal frequency, which represents oscillations per second.
To put it simply, while temporal frequency counts the number of cycles per second, angular frequency counts the angle traversed in radians per second. For a standard cosine function \( \, \cos(\omega t) \), \(\omega\) defines how fast the cosine wave swings back and forth.
Period
The period (\(T\)) of a cosinusoidal signal is the time it takes to complete one full cycle of oscillation. It is inversely related to the temporal frequency \(f\) by: \[ T = \frac{1}{f} \]
Understanding the period is critical as it tells us how long one complete oscillation lasts. Here's how to connect this with our cosinusoidal signal formula:
  • A longer period means a slower oscillating signal.
  • A shorter period indicates a faster oscillation.

Since \(T\) is the duration of one cycle, you can see that in each second, \(f\) cycles occur—matching the definition that \(f\) is the number of cycles per second. This relationship ensures that knowing either the period or the frequency helps in comprehending the tempo of oscillations in the signal.

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Most popular questions from this chapter

Show that $$ \operatorname{Re}\left[V e^{j \omega t}\right]=A \cos (\omega t+\varphi) $$ when \(V=A e^{j \varphi}\).

Prove that the operators \(\frac{d}{d t}\) and \(\mathrm{Re}[]\) commute: \(\frac{d}{d t} \operatorname{Re}\left\\{e^{j \omega t}\right\\}=\operatorname{Re}\left\\{\frac{d}{d t} e^{j \omega t}\right\\}\)

Write the phasor \(A_{1} e^{j \varphi_{1}}+A_{2} e^{j \varphi_{2}}\) as \(A_{3} e^{j \varphi_{3}} ;\) determine \(A_{3}\) and \(\varphi_{3}\) in terms of \(A_{1}, A_{2}, \varphi_{1}\), and \(\varphi_{2} .\) What is the corresponding real signal?

Show how to compute \(A\) and \(\varphi\) in the equation \(x(t)=A \cos (\omega t+\varphi)\) from the initial conditions \(x(O)\) and \(\left.\frac{d}{d t} x(t)\right|_{t=0}\).

Sketch the imaginary part of \(A e^{j \varphi} e^{j \omega t}\) to show that this is \(A \sin (\omega t+\varphi)\). What do we mean when we say that the real and imaginary parts of \(A e^{j \varphi} e^{j \omega t}\) are " \(90^{\circ}\) out of phase"?
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