DISCRETE-TIME SINUSOIDAL SIGNALS x[n]=Acos(Ωn+Φ); −∞<n<∞ n is integer (discrete-time variable) n is called the sample number. A: amplitude Ω: discrete frequency (radians/sample( Φ: phase (radians) Instead of Ω, we can use the discrete cyclic frequency variable, F. Ω=2πF ⟹x n =Acos 2πFn+∅ ; −∞<n<∞ F: discrete cyclic frequency (cycles/sample) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

x[n]=Acos π 6 n+ π 3 ⟹Ω= π 6 rad/samples Example: x[n]=Acos π 6 n+ π 3 ⟹Ω= π 6 rad/samples ⟹F=⋯ cycles sample and ∅= π 3 radians Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Properties of Discrete-Time Sinusoids: Property 1- A discrete-time sinusoid is periodic ONLY IF ITS FREQUENCY F IS RATIONAL NUMBER I (Compare with continuous sinusoids: A continuous-time sinusoid is always periodic, for any value of its frequency, f0) Definition: x[n] is periodic with period N (N>0 and integer) if and only if x[n]=x[n+N] for all n. The smallest value of N for which above equation is true is called the fundamental period. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Proof of Periodicity Property: If x[n] is periodic; then cos 2πF N+n +∅ x[n+N] = cos 2πFn+∅ x[n] This equation is true if there exists an integer K such that 2πFN=k2π ⟹F= k N which is a rational number. Thus, x[n] is periodic if and only if its frequency F can be expressed as the ration of two integers. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

To determine the fundamental period N of a periodic discrete-time sinusoid, we express its frequency F as a rational number, F= k N , and cancel common factors so that K and N are relatively prime. Then, the fundamental period is equal to N. For example, If F 1 = 31 60 then N 1 =60 samples (sinusoid repeats itself every 60 samples) If F 1 = 30 60 then N 2 =2 samples (sinusoid repeats itself every 2 samples) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

cos Ω+2π n+∅ =cos Ωn+2πn+∅ =cos Ωn+∅ Property 2- Discrete-time sinusoids whose frequencies are separated by an integer multiple of 2π are identical. cos Ω+2π n+∅ =cos Ωn+2πn+∅ =cos Ωn+∅ Thus, any two discrete-time sinusoids with frequencies in the range −π≤Ω≤π or − 1 2 ≤F≤ 1 2 are distinct (unique). A sinusoid with Ω >π (or F > 1 2 ) is identical to a sinusoid with frequency Ω <π (or F < 1 2 ) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

-∞< ω 0 <∞ or −∞< f 0 <∞) We call the sinusoid with Ω >π , an alias of a corresponding sinusoid with Ω <π . (Compare with continuous sinusoids: Continuous-time sinusoids are distinct for all frequency values -∞< ω 0 <∞ or −∞< f 0 <∞) Property 3- Highest rate of oscillation in a discrete-time sinusoid is attained when Ω=π (or Ω=−π) (or equivalently, for F=1/2 or -1/2) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

To illustrate this property, let x[n]=cos (Ω1n) Consider frequency Ω1n=0, π/8, π/4, π/2, π (or equivalently, F1=0, 1/16, 1/8, 1/4, 1/2) which result in periodic discrete-time sinusoids with periods N= ∞, 16, 8, 4, 2. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

x 2 n =Acos 2π −Ω 1 n =Acos Ω 1 n = x 1 n To see what happens for π<Ω<2π, we consider the sinusoids with Ω1 and 2π-Ω1. Note that as Ω1 varies from π to 2π, 2π-Ω1 varies from π to 0. Let x 1 n =Acos Ω 1 n x 2 n =Acos 2π −Ω 1 n =Acos Ω 1 n = x 1 n Thus, x2[n] is an alias of x1[n]. As with continuous-time sinusoids, complex exponentials and negative frequencies can be introduced for discrete-time sinusoids. x n =Acos Ωn+ϕ = A 2 e j(Ωn+ϕ) + A 2 e −j(Ωn+ϕ) (Euler formula) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

which we call the fundamental range. Since discrete-time sinusoids with frequencies that are seperated by an integer multiple of 2π are identical, frequencies in any interval Ω 1 ≤Ω≤ Ω 1 +2π constitute all the existing discrete-time sinusoids or complex exponentials. Hence, frequency range for discrete-time sinusoids is finite with duration 2π. Usually, we choose the range 0≤Ω≤2π or −π≤Ω≤π (0≤F≤1 or − 1 2 ≤F≤ 1 2 ) which we call the fundamental range. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

SAMPLING OR ANALOG SIGNALS Sampling is described by the equation x n = x a n T s ; −∞<n<∞ x[n] is the discrete-time signal obtained by taking samples of the analog signal xa(t) every Ts seconds. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Relation between continuous-time, t and discrete-time, n: t=n T s = n f s As a result of this, there exists a relationship between continuous frequency f0 (ω0) and discrete frequency F (or Ω). Consider analog sinusoid xa(t)=Acos(2πf0nTs+Φ) sample with T S ⟹ x a n T S =x n =Acos 2π f 0 n T S +ϕ =Acos( 2π f 0 n f S +ϕ) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

−∞< f 0 <∞ and −∞< ω 0 <∞ ⟹discrete frequency F= f 0 f S or equivalently Ω= ω 0 T S Recall that for continuous-time −∞< f 0 <∞ and −∞< ω 0 <∞ However, for discrete-time sinusoids −1 2 ≤F≤ 1 2 and ⇓ −1 2 ≤ f 0 f S ≤ 1 2 ⇓ −1 2 T S = − f S 2 < f 0 < f S 2 = 1 T S −π≤Ω≤π ⇓ −π≤ ω 0 T S ≤π ⇓ −π T S =−π f S ≤ ω 0 ≤π f S = π T S Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

f 0 max= f S 2 = 1 2 T S or ω 0 max=π f S = π T S Since the highest frequency in a discrete-time signal is Ω=π (or F= 1 2 ), It follows that, with sampling rate fS, the corresponding highest values for f0 and ω0 are f 0 max= f S 2 = 1 2 T S or ω 0 max=π f S = π T S What happens to frequencies above f S 2 ? Consider following example: Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Let x1(t) and x2(t) be sampled with sampling frequency fs=40 Hz. Ex: x1(t)=cos(2π10t) x2(t)=cos(2π50t) Let x1(t) and x2(t) be sampled with sampling frequency fs=40 Hz. x 1 n = cos 2π 10 40 n =cos⁡( π 2 n) x 2 n = cos 2π 50 40 n =cos⁡( 5π 2 n) However, since cos 5π 2 n =cos 2πn+ π 2 n =cos⁡( π 2 n) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

⟹ x 1 n = x 2 [n]⟹discrete sinusoidal signals are identical. Thus, we say that the frequency f02=50 Hz is an alias of the frequency f01=10 Hz at the sampling rate of fS=40 samples/sec. f02=50 Hz is not the only alias of f01=10 Hz. In fact, for fS=40 samples/sec, f03=90 Hz is also an alias. f04=130 Hz is also an alias. All of the continuous sinusoids cos[2π(10+40k)t]; k=1, 2, 3, 4, ... when sampled at 40 samples/sec, give identical values. ⟹they are all aliases of fo1=10 Hz. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

In general, sampling continuous sinusoid xa(t)=Acos(2πf0nTs+Φ) with sampling rate f s = 1 T s Results in the discrete sinusoid x[t]=Acos(2πFn+Φ) where F= f 0 T s If we assume − f S 2 < f 0 < f S 2 , frequency F of x[n] is the range −1 2 ≤F≤ 1 2 which is the range for discrete sinusoids Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

On the other hand, if the sinusoids xa(t)=Acos(2πf0kt+Φ) where f0k=f0+kfS Are sampled at fS; then the sampled (discrete) sinusoid x n = x a n T S =Acos(2π f 0 +k f S f S n+∅) =Acos 2πn f 0 f S F +∅+2πkn =Acos(2πFn+∅) which is identical to the discrete-time sinusoid obtained by sampling Acos(2πf0t+Φ) with fS. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Thus, frquencies f0k=f0+kfS; k=∓1, ∓2, ⋯ Are indistinguishable from the frequency f0 after sampling with fS; hence they are aliases of f0. Example of aliasing: Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Two continuous-time with f 01 = 1 8 Hz, f 02 =− 7 8 Hz, sampling rate fS=1 Hz (k=+1) Note that for k=+1 f01=f02+fS=(-7/8+1)=1/8 The frequency f0=fS/2 is called the folding frequency. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Copyright 2012 | Instructor: Dr Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Example: Consider the analog signal xa(t)=3cos(100πt) Determine the minimum sampling rate to avoid aliasing. Answer: since f0=50 Hz, the min. sampling rate to avoid aliasing is fS=100 Hz. (b) If fS=200 Hz, what is the discrete-time signal obtained after sampling? Answer: x n =3 cos 100πn 200 =3cos⁡( π 2 n) Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

x n =3 cos 100πn 75 =3 cos 4π 3 n =3 cos 2π− 2π 3 n =3cos⁡( 2π 3 n) (c) If fS=75 Hz, what is the discrete-time signal obtained after sampling? Answer: x n =3 cos 100πn 75 =3 cos 4π 3 n =3 cos 2π− 2π 3 n =3cos⁡( 2π 3 n) (d) What is the frequency 0<f0<fS/2 of a sinusoid that gives samples identical to those obtained in part (c)? Answer: for fS=75 Hz we have f0=FfS=75F Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

⟹f0=50 Hz is an alias of f0=25 Hz for fS=75 Hz. in part (c), we found the discrete-frequency F=1/3. Hence, f 0 =75. 1 3 =25 Hz. ⟹ the sinusoid ya(t)=3cos(2πf0t)=3cos(50πt) sampled at fS=75 samples/sec gives identical samples. ⟹f0=50 Hz is an alias of f0=25 Hz for fS=75 Hz. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

The Sampling Theorem Given an analog signal, how should we select the sampling period TS, or equivalently, the sampling rate, fS? We know that the highest frequency in an analog signal that can be unambiguously reconstructed when the signal is sampled at a rate fS=1/Ts is fS/2. Any frequency above fS/2 or below –fS/2 results in samples that are identical with a corresponding frequency in the range − f S 2 < f 0 < f S 2 . To avoid the ambiguities resulting from aliasing, we must select the sampling rate to be sufficiently high. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

f S >2 f 0 max (sampling theorem) That is, we must select the fS/2˃f0max where f0max is the highest possible frequency of the analog signal. Thus, to avoid the problem of aliasing, fS is selected so that f S >2 f 0 max (sampling theorem) With the sampling rate selected in this manner, any frequency component f 0 < f 0 max in the analog signal is mapped into a discrete-time frequency value −1 2 ≤F= f 0 f S ≤ 1 2 or equivalently −π≤Ω=2πF≤π The sampling rate fS=2f0max is called the Nyquist rate. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING

Example: Consider the analog signal xa(t)=3cos(50πt)+10sin(300πt)-cos(100πt) What is the Nyquist rate for this signal? Answer: frequencies present in xa(t) are f1=25 Hz, f2=150 Hz, f3=50 Hz ⟹f0max=150 Hz ⟹Nyquist rate=300 Hz We should satisfy f 0 >300 samples/sec to avoid aliasing. Copyright 2012 | Instructor: Dr. Gülden Köktürk | EED1004-INTRODUCTION TO SIGNAL PROCESSING