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Signals & Systems

Lecture 3 – Spectrum and Fourier Series

Alp Ertürk

alp.erturk@kocaeli.edu.tr

Sinusoidal Signals:Phasor Addition

• The following statement is satisfied when we add several sinusoidal signals with the same frequency, but with different amplitudes and phases

𝑘=1

𝑁

𝐴𝑘 cos 𝜔0𝑡 + 𝜙𝑘 = 𝐵 cos 𝜔0𝑡 + 𝜃

• In other words, we obtain a single sinusoidal signal with the same frequency.

Spectrum

• What if the frequencies are different too?

𝑘=1

𝑁

𝐴𝑘 cos 2𝜋𝑓𝑘𝑡 + 𝜙𝑘 = ?

• Or, similarly:

𝑥 𝑡 = 𝐴0 +

𝑘=1

𝑁

𝐴𝑘 cos 2𝜋𝑓𝑘𝑡 + 𝜙𝑘 = ?

Spectrum

𝑥 𝑡 = 10 + 14 cos 200𝜋𝑡 − 𝜋 3 + 8 cos 500𝜋𝑡 + 𝜋 2 =?

• Using Euler’s formula of:

𝐴 cos 2𝜋𝑓𝑡 + 𝜙 =𝐴

2𝑒𝑗𝜙𝑒𝑗2𝜋𝑓𝑡 +

𝐴

2𝑒−𝑗𝜙𝑒−𝑗2𝜋𝑓𝑡

• The first sinusoidal signal is:

14 cos 200𝜋𝑡 − 𝜋 3 =14

2𝑒𝑗 − 𝜋 3 𝑒𝑗2𝜋 100 𝑡 +

14

2𝑒−𝑗 − 𝜋 3 𝑒−𝑗2𝜋 100 𝑡

Spectrum

𝑥 𝑡 = 10 + 14 cos 200𝜋𝑡 − 𝜋 3 + 8 cos 500𝜋𝑡 + 𝜋 2 =?

• We get:

𝑥 𝑡= 10 + 7𝑒−𝑗 𝜋 3𝑒𝑗2𝜋 100 𝑡 + 7𝑒𝑗 𝜋 3𝑒−𝑗2𝜋 100 𝑡

+ 4𝑒𝑗 𝜋 2𝑒𝑗2𝜋 250 𝑡 + 4𝑒−𝑗 𝜋 2𝑒−𝑗2𝜋 250 𝑡

Spectrum

• In other words:

𝑥 𝑡 = 𝐴0 +

𝑘=1

𝑁

𝐴𝑘 cos 2𝜋𝑓𝑘𝑡 + 𝜙𝑘

=

𝑘=−𝑁

𝑁

𝑎𝑘𝑒𝑗2𝜋𝑓𝑘𝑡

• where

𝑎𝑘 =

𝐴0 𝑘 = 01

2𝐴𝑘𝑒

𝑗𝜙𝑘 𝑘 ≠ 0

Spectrum

𝑥 𝑡 = 10 + 14 cos 200𝜋𝑡 − 𝜋 3 + 8 cos 500𝜋𝑡 + 𝜋 2 =?

= 10 + 7𝑒−𝑗 𝜋 3𝑒𝑗2𝜋 100 𝑡 + 7𝑒𝑗 𝜋 3𝑒−𝑗2𝜋 100 𝑡

+ 4𝑒𝑗 𝜋 2𝑒𝑗2𝜋 250 𝑡 + 4𝑒−𝑗 𝜋 2𝑒−𝑗2𝜋 250 𝑡

Spectrum

• The above plot is named as the spectrum plot

• If a signal is composed of a finite number of sinusoidal components, we can find its spectral components

Spectrum

• Another representation:

0 , 10 , 100 , 7𝑒−𝑗 𝜋 3 , −100 , 7𝑒𝑗 𝜋 3 , 250 , 4𝑒𝑗 𝜋 2 , −250 , 4𝑒−𝑗 𝜋 2

Spectrum

• What if we have sinusoidal multiplication instead of summation?

Spectrum• If we have sinusoidal multiplication instead of addition, then

we have to rewrite the multiplication as an additive combination of sinusoidals

• For example:

𝑥 𝑡 = cos(𝜋𝑡) sin 10𝜋𝑡

• Represent as:

𝑥 𝑡 =𝑒𝑗𝜋𝑡 + 𝑒−𝑗𝜋𝑡

2

𝑒𝑗10𝜋𝑡 − 𝑒−𝑗10𝜋𝑡

2𝑗

Spectrum• For example:

𝑥 𝑡 = cos(𝜋𝑡) sin 10𝜋𝑡

• Represent as:

𝑡 =𝑒𝑗𝜋𝑡 + 𝑒−𝑗𝜋𝑡

2

𝑒𝑗10𝜋𝑡 − 𝑒−𝑗10𝜋𝑡

2𝑗

=1

4𝑒−𝑗 𝜋 2𝑒𝑗11𝜋𝑡 +

1

4𝑒−𝑗 𝜋 2𝑒𝑗9𝜋𝑡 −

1

4𝑒−𝑗 𝜋 2𝑒−𝑗9𝜋𝑡 −

1

4𝑒−𝑗 𝜋 2𝑒−𝑗11𝜋𝑡

=1

4𝑒−𝑗 𝜋 2𝑒𝑗11𝜋𝑡 +

1

4𝑒−𝑗 𝜋 2𝑒𝑗9𝜋𝑡 +

1

4𝑒𝑗 𝜋 2𝑒−𝑗9𝜋𝑡 +

1

4𝑒𝑗 𝜋 2𝑒−𝑗11𝜋𝑡

=1

2cos 11𝜋𝑡 − 𝜋 2 +

1

2cos 9𝜋𝑡 − 𝜋 2

Spectrum

𝑥 𝑡 = cos(𝜋𝑡) sin 10𝜋𝑡

=1

2cos 11𝜋𝑡 − 𝜋 2 +

1

2cos 9𝜋𝑡 − 𝜋 2

=1

4𝑒−𝑗 𝜋 2𝑒𝑗11𝜋𝑡 +

1

4𝑒𝑗 𝜋 2𝑒−𝑗11𝜋𝑡 +

1

4𝑒−𝑗 𝜋 2𝑒𝑗9𝜋𝑡 +

1

4𝑒𝑗 𝜋 2𝑒−𝑗9𝜋𝑡

Spectrum

• How does the multiplication of two sinusoidals appear in time domain?

𝑥 𝑡 = 2 cos(2𝜋 20 𝑡) cos 2𝜋 200 𝑡

• The two components individually:

Spectrum

• How does the multiplication of two sinusoidals appear in time domain?

𝑥 𝑡 = 2 cos(2𝜋 20 𝑡) cos 2𝜋 200 𝑡

• The two components multiplied:

Spectrum

• MATLAB Visual Example

Spectrum

• Matlab code:clear all; close all; clc;

t = 0:0.0001:0.1;

y1 = 2*cos(2*pi*20*t);

y2 = 2*cos(2*pi*200*t);

y3 = y1.*y2/2;

figure; plot(t,y1,'LineWidth',2);

legend('2*cos(2*pi*20*t)'); xlabel('time (s)'); ylabel('Amplitude');

pause; hold on; plot(t,y2,'r','LineWidth',2);

legend('2*cos(2*pi*20*t)' , '2*cos(2*pi*200*t)');

pause; hold on; plot(t,-y1,'b-.','LineWidth',2);

legend('2*cos(2*pi*20*t)' , '2*cos(2*pi*200*t)', '-2*cos(2*pi*20*t)');

pause; hold on; plot(t,y3,'k','LineWidth',2);

legend('2*cos(2*pi*20*t)' , '2*cos(2*pi*200*t)', '-2*cos(2*pi*20*t)' , '2*cos(2*pi*20*t)*cos(2*pi*200*t)');

pause; figure; plot(t,y3,'k','LineWidth',2);

xlabel('time (s)'); ylabel('Amplitude'); title('2*cos(2*pi*20*t)*cos(2*pi*200*t)');

Spectrum

• Multiplying sinuoidals is used in communication systems for modulation

• Amplitude modulation (AM) is the process of multiplying a low-frequency message signal by a high-frequency carrier signal

• AM can be represented by:

𝑥 𝑡 = 𝑣(𝑡) cos 2𝜋𝑓𝑐𝑡

where v(t) is the message signal, and the sinusoidal with the fc frequency is the carrier signal

Spectrum

• MATLAB Visual Example

Spectrum

• t = 0:0.0001:0.1;

• v = 5 + 4 * cos(2*pi*20*t);

• c = cos(2*pi*200*t);

• x = v .* c;

•

• figure; plot(t,v);

• hold on; plot(t,-v,'-.');

• hold on; plot(t,c,'r');

• hold on; plot(t,x,'g');

• xlabel('time (s)'); ylabel('Amplitude');

• legend('Message' , ' - Message' , 'Carrier' , 'AM Signal');

Spectrum

• Let v(t) = 5 + 4 cos (40πt) and fc = 200 Hz

𝑥 𝑡 = 5 + 4 cos 40𝜋𝑡 cos 400𝜋𝑡

Spectrum

• The term amplitude modulation (AM) is used because the effect of multiplying the higher-frequency sinusoid by the lower frequency sinusoid is to ‘‘modulate’’ the amplitude envelope of the carrier waveform

Spectrum

• Let v(t) = 5 + 4 cos (40πt) and fc = 200 Hz

𝑥 𝑡 = 5 + 4 cos 40𝜋𝑡 cos 400𝜋𝑡

=5

2𝑒𝑗400𝜋𝑡 + 𝑒𝑗440𝜋𝑡 + 𝑒𝑗360𝜋𝑡 +

5

2𝑒−𝑗400𝜋𝑡 + 𝑒−𝑗440𝜋𝑡 + 𝑒−𝑗360𝜋𝑡

Spectrum

• We had mentioned periodic signals and fundamental frequency before

• Can we synthesize a periodic signal by adding two or more cosine waves?

• Yes, if the cosine waves have harmonically related frequencies

• Harmonically related frequencies: Frequencies that are integer multiples of a frequency

Spectrum

• MATLAB Visual Example

Spectrum

• t = 0:0.01:1;

• s1 = cos(2*pi*1.2*t);

• s2 = cos(2*pi*2*t);

• s3 = cos(2*pi*6*t);

• figure; plot(t,s1);

• hold on; plot(t,s2,'--');

• hold on; plot(t,s3,'-.');

• hold on; plot(t,s1+s2+s3,'r');

• legend('cos(2*pi*1.2*t)' , 'cos(2*pi*2*t)' , 'cos(2*pi*6*t)', 'Sum of all three');

• xlabel('time (s)'); ylabel('Amplitude'); t = 0:0.01:5;

• s1 = cos(2*pi*1.2*t);

• s2 = cos(2*pi*2*t);

• s3 = cos(2*pi*6*t);

• figure; plot(t,s1+s2+s3,'r');

• xlabel('time (s)'); ylabel('Amplitude');

Spectrum

𝑥 𝑡 = cos 2𝜋(1.2)𝑡 + cos 2𝜋(2)𝑡 + cos 2𝜋(6)𝑡

• The fundamental frequency of x(t) is 0.4 Hz

Spectrum

Spectrum

Spectrum

• The vowel sound ‘‘ah’’ can be approximated by the following complex amplitudes

• The spectrum obtained by adding these signals is the addition of their respective spectrums:

Spectrum

• In time domain, adding the signals up one at a time:

Spectrum

Spectrum

• Hence the total signal has a fundamental frequency of 100 Hz

Spectrum

• We can synthesize a periodic signal by adding two or more cosine waves if the cosine waves have harmonically related frequencies

Spectrum

𝑥 𝑡 = 2 cos 20𝜋𝑡 −2

3cos 20𝜋(3)𝑡 +

2

3cos 20𝜋(5)𝑡

Spectrum

𝑥 𝑡 = 2 cos 20𝜋𝑡 −2

3cos 20𝜋(3)𝑡 +

2

3cos 20𝜋(5)𝑡

Spectrum

• What if the frequencies of the added sinusoidal signals are not harmonically related?

Spectrum

𝑥 𝑡 = 2 cos 20𝜋𝑡 −2

3cos 20𝜋( 8)𝑡 +

2

3cos 20𝜋( 27)𝑡

Spectrum

𝑥 𝑡 = 2 cos 20𝜋𝑡 −2

3cos 20𝜋( 8)𝑡 +

2

3cos 20𝜋( 27)𝑡

Fourier Series

• Any periodic signal can be synthesized with a sum of harmonically related sinusoids *

• Note that this may require an infinite number of terms

𝑥 𝑡 =

𝑘=−∞

∞

𝑎𝑘𝑒𝑗 2𝜋 𝑇0 𝑘𝑡

* This is not 100% correct, and we will get back to that point later on

Fourier Series

𝑥 𝑡 =

𝑘=−∞

∞

𝑎𝑘𝑒𝑗 2𝜋 𝑇0 𝑘𝑡

• When the complex amplitudes are conjugate-symmetric, i.e. 𝑎−𝑘 = 𝑎𝑘

∗ , the synthesis formula becomes:

𝑥 𝑡 = 𝐴0 +

𝑘=1

∞

𝐴𝑘 cos 2𝜋 𝑇0 𝑘𝑡 + 𝜙𝑘

Fourier Series

• Driving the coefficients for the harmonic sum of a periodic signal:

𝑎𝑘 =1

𝑇0

0

𝑇0

𝑥(𝑡)𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡

• As a special case, the DC component is obtained by:

𝑎0 =1

𝑇0

0

𝑇0

𝑥 𝑡 𝑑𝑡

Fourier Series: Example

• Determine the Fourier series coefficients of 𝑥 𝑡 = sin3 3𝜋𝑡

Fourier Series: Example

• Determine the Fourier series coefficients of 𝑥 𝑡 = sin3 3𝜋𝑡

• In this case, an easier approach is to use Euler’s formula

𝑥 𝑡 = sin3 3𝜋𝑡 =𝑒𝑗3𝜋𝑡 − 𝑒−𝑗3𝜋𝑡

2𝑗

3

=1

−8𝑗𝑒𝑗9𝜋𝑡 − 3𝑒𝑗6𝜋𝑡𝑒−𝑗3𝜋𝑡 + 3𝑒𝑗3𝜋𝑡𝑒−𝑗6𝜋𝑡 − 𝑒−𝑗9𝜋𝑡

=𝑗

8𝑒𝑗9𝜋𝑡 +

−3𝑗

8𝑒𝑗3𝜋𝑡 +

3𝑗

8𝑒−𝑗3𝜋𝑡 +

−𝑗

8𝑒−𝑗9𝜋𝑡

Fourier Series: Example

𝑥 𝑡 = sin3 3𝜋𝑡 =𝑒𝑗3𝜋𝑡 − 𝑒−𝑗3𝜋𝑡

2𝑗

3

=𝑗

8𝑒𝑗9𝜋𝑡 +

−3𝑗

8𝑒𝑗3𝜋𝑡 +

3𝑗

8𝑒−𝑗3𝜋𝑡 +

−𝑗

8𝑒−𝑗9𝜋𝑡

• Therefore,𝜔0 = 3𝜋

𝑎𝑘 =∓𝑗

3

8𝑘 = ±1

±𝑗1

8𝑘 = ±3

Fourier Series: Example

𝑥 𝑡 =𝑗

8𝑒𝑗9𝜋𝑡 +

−3𝑗

8𝑒𝑗3𝜋𝑡 +

3𝑗

8𝑒−𝑗3𝜋𝑡 +

−𝑗

8𝑒−𝑗9𝜋𝑡

𝜔0 = 3𝜋 , 𝑎𝑘 =∓𝑗

3

8𝑘 = ±1

±𝑗1

8𝑘 = ±3

Fourier Series: Another Example

• Determine the Fourier series coefficients of:

𝑥 𝑡 = 1 + sin𝜔0𝑡 + 2 cos𝜔0𝑡 + cos 2𝜔0𝑡 +𝜋

4

Fourier Series: Another Example

𝑥 𝑡 = 1 + sin𝜔0𝑡 + 2 cos𝜔0𝑡 + cos 2𝜔0𝑡 +𝜋

4

= 1 +1

2𝑗𝑒𝑗𝜔0𝑡 − 𝑒−𝑗𝜔0𝑡 + 1 𝑒𝑗𝜔0𝑡 + 𝑒−𝑗𝜔0𝑡

+1

2𝑒𝑗 2𝜔0𝑡+ 𝜋 4 + 𝑒−𝑗 2𝜔0𝑡+ 𝜋 4

• Collecting terms:

= 1 + 1 +1

2𝑗𝑒𝑗𝜔0𝑡 + 1 −

1

2𝑗𝑒−𝑗𝜔0𝑡 +

1

2𝑒𝑗𝜋/4 𝑒𝑗2𝜔0𝑡

+1

2𝑒−𝑗𝜋/4 𝑒−𝑗2𝜔0𝑡

Fourier Series: Another Example

𝑎0 = 1

𝑎1 = 1 +1

2𝑗= 1 −

1

2𝑗

𝑎−1 = 1 −1

2𝑗= 1 +

1

2𝑗

𝑎2 =1

2𝑒𝑗𝜋/4 =

2

41 + 𝑗

𝑎−2 =1

2𝑒−𝑗𝜋/4 =

2

41 − 𝑗

𝑎𝑘 = 0 , 𝑘 > 2

Fourier Series: Square Wave

• Consider the periodic square wave, which is defined for one cycle by:

𝑠 𝑡 =1 for 0 ≤ 𝑡 <

1

2𝑇0

0 for1

2𝑇0 ≤ 𝑡 < 𝑇0

Fourier Series: Square Wave

𝑎𝑘 =1

𝑇0

0

𝑇0

𝑥(𝑡)𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡

=1

𝑇0

0

𝑇0/2

1𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡

=1

𝑇0

𝑒−𝑗 2𝜋 𝑇0 𝑘(

12𝑇0) − 𝑒−𝑗 2𝜋 𝑇0 𝑘(0)

−𝑗 2𝜋 𝑇0 𝑘

Fourier Series: Square Wave

=1

𝑇0

𝑒−𝑗 2𝜋 𝑇0 𝑘(

12𝑇0) − 𝑒−𝑗 2𝜋 𝑇0 𝑘(0)

−𝑗 2𝜋 𝑇0 𝑘

=𝑒−𝑗𝜋𝑘 − 1

−𝑗2𝜋𝑘

𝑎𝑘 =−1 𝑘 − 1

−𝑗2𝜋𝑘, for 𝑘 ≠ 0

𝑎0 =1

𝑇0

0

𝑇0/2

1𝑒−𝑗0𝑡𝑑𝑡 =1

𝑇0

1

2𝑇0 =

1

2

Fourier Series: Square Wave

𝑠 𝑡 =1 for 0 ≤ 𝑡 <

1

2𝑇0

0 for1

2𝑇0 ≤ 𝑡 < 𝑇0

𝑎𝑘 =

1

𝑗𝜋𝑘𝑘 = ±1,±3,±5,…

0 𝑘 = ±2,±4,±6,…1

2𝑘 = 0

Fourier Series: Square Wave

• For the 50% duty cycle square wave, if the fundamental frequency is 25 Hz:

Fourier Series: Triangle Wave

• Consider the periodic square wave, which is defined for one cycle by:

𝑠 𝑡 = 2𝑡 𝑇0 for 0 ≤ 𝑡 <

1

2𝑇0

2 𝑇0 − 𝑡 𝑇0 for1

2𝑇0 ≤ 𝑡 < 𝑇0

• where 𝑇0 = 0.04 s

Fourier Series: Triangle Wave

𝑎0 =1

𝑇0

0

𝑇0

𝑥 𝑡 𝑑𝑡 =1

𝑇0𝐴𝑟𝑒𝑎 =

1

2

𝑎𝑘 =1

𝑇0

0

𝑇0

𝑥(𝑡)𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡

=1

𝑇0

0

𝑇0/2

( 2𝑡 𝑇0)𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡 +

1

𝑇0

𝑇0/2

𝑇0

( 2 𝑇0 − 𝑡 𝑇0)𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡

Fourier Series: Triangle Wave

• After some steps, we arrive at:

𝑎𝑘 =𝑒−𝑗𝑘𝜋 − 1

𝜋2𝑘2

• Since 𝑒−𝑗𝑘𝜋 = (−1)𝑘 :

𝑎𝑘 =

−2

𝜋2𝑘2𝑘 = ±1,±3,±5,…

0 𝑘 = ±2,±4,±6,…1

2𝑘 = 0

Convergence of Fourier Series

• Some History:

• Euler and Lagrange would agree with our examples before the square wave, but have objected to the square wave example

• However, Fourier maintained that the Fourier representation of the square wave is valid

• In fact, Fourier maintained that any periodic signal could be represented by a Fourier series

Convergence of Fourier Series

• This is not quite correct

• However, an extremely large class of periodic signals, including the square wave, can be reprented by a Fourier series

• To illuminate this point, let us consider approximating a given periodic signal by a linear combination of a finite number of harmonically related complex exponentials

Convergence of Fourier Series

𝑥𝑁 𝑡 =

𝑘=−𝑁

𝑁

𝑎𝑘𝑒𝑗𝑘𝜔0𝑡

• Then, an approximation error can be calculated by:

𝑒𝑁 𝑡 = 𝑥 𝑡 − 𝑥𝑁 𝑡 = 𝑥 𝑡 −

𝑘=−𝑁

𝑁

𝑎𝑘𝑒𝑗𝑘𝜔0𝑡

Convergence of Fourier Series

• Error magnitudes when approaching a square wave with a sum of harmonic components:

Convergence of Fourier Series

• To derive a quantitative measure of approximation error, we can use the energy of the error signal over one period:

𝐸𝑁 = 𝑇

𝑒𝑁(𝑡)2𝑑𝑡

• To minimize this energy:

𝑎𝑘 =1

𝑇0

0

𝑇0

𝑥(𝑡)𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡

Convergence of Fourier Series

𝐸𝑁 = 𝑇

𝑒𝑁(𝑡)2𝑑𝑡

𝑎𝑘 =1

𝑇0

0

𝑇0

𝑥(𝑡)𝑒−𝑗 2𝜋 𝑇0 𝑘𝑡𝑑𝑡

• If a signal x(t) has a Fourier series representation, limit of 𝐸𝑁as N →∞ is zero

• What if the integral diverges?

Convergence of Fourier Series

• Three conditions, derived by Dirichlet, guarantees that a given signal equals its Fourier series representation

• , except at isolated values of t for which x(t) is discontinuous

Convergence of Fourier Series

Condition 1:

Over any period, x(t) must be absolutely integrable:

𝑇

𝑥(𝑡) 𝑑𝑡 < ∞

This guarantees that each coefficient 𝑎𝑘 will be finite

Convergence of Fourier Series

• A periodic signal that violates the first Dirichlet condition is:

𝑥 𝑡 =1

𝑡, 0 < 𝑡 ≤ 1

Convergence of Fourier Series

Condition 2:

In any finite interval of time, x(t) is of bounded variation

In other words, there are no more than a finite number of maxima and minima during any single period of the signal

Convergence of Fourier Series

• A periodic signal that satisfies the first Dirichlet condition, but violates the second condition is:

𝑥 𝑡 = sin2𝜋

𝑡, 0 < 𝑡 ≤ 1

Convergence of Fourier Series

Condition 3:

In any finite interval of time, there are only a finite number of discontinuities.

Furthermore, each of these discontinuities is finite

Convergence of Fourier Series

A signal that violates this condition:

Frequency Variation

• MATLAB Audio Example

Frequency Variation

• A simple case of signals with varying frequency is the chirp signal

• A chirp signal’s frequency changes linearly from some low value to a high value

• A simple way to obtain chirp signals would be concatenate a number of sinusoids with different constant frequencies

• But this causes discontinuities in the boundaries unless the initial phases are carefully adjusted

Frequency Variation

Frequency Variation

• A better appraoch is to modify the sinusoid signal formula

𝑥 𝑡 = 𝐴 cos 𝜓(𝑡)

• As an example,

𝑥 𝑡 = 𝐴 cos 2𝜋𝜇𝑡2 + 2𝜋𝑓0𝑡 + 𝜙

𝑓𝑖 𝑡 =1

2𝜋

𝑑𝜓(𝑡)

𝑑𝑡= 2𝜇𝑡 + 𝑓0

Frequency Variation

• Suppose we want to synthesize a frequency sweep from f1 = 220Hz to f2 = 2320Hz over a 3 – second time interval. Then,

𝑓𝑖 𝑡 =𝑓2 − 𝑓1𝑇2

𝑡 + 𝑓1 =2320 − 220

3𝑡 + 220

𝜓 𝑡 =

0

𝑡

2𝜋𝑓 𝜏 𝑑𝜏 =

0

𝑡

2𝜋2320 − 220

3𝜏 + 220 𝑑𝜏

= 700𝜋𝑡2 + 440𝜋𝑡 + 𝜙

• Note: A frequency variation produced by the time-varying angle function is called frequency modulation (FM)