Concept of frequency in continuous-time and discrete-time - Digital Signal Processing, Engineering - Computer Science Engineering (CSE) PDF Download

Concept of frequency in continuous-time and discrete-time.

1)  x_a (t)   = A Cos ( Ω t)

x (nTs) = A Cos ( Ω nTs)

= A Cos (wn)

w = Ω Ts

Ω = rad / sec           w = rad / Sample  

F = cycles / sec      f = cycles / Sample 

2) A Discrete- time – sinusoid is periodic only of its f is a Rational number.

x (n+N) = x (n)

Cos 2π f₀ (n+N) = Cos 2 π f₀ n  

2 π f₀N = 2π K =>   f₀ =K/N

Ex:   A Cos (π/6)n

f=1/12  N=12 Samples/Cycle ;   Fs= Sampling Frequency; Ts =

Sampling Period

Q. Cos (0.5n) is not periodic

Q. x (n) = 5 Sin (2n)

2π f = 2   => f =1/π Non-periodic

Q. x (n) = 5 Cos (6π n)

2π f = 6π => f = 3                N=1 for K=3   Periodic

Q. x (n) = 5 Cos 6πn/35

2π f = 6π/35 =>f=3/35         for N=35 & K=3       Periodic    

Q. x (n) = Sin (0.01π n)    

2π f = 0.01π => f =0.01/2       for N=200 & K=1       Periodic

Q. x (n) = Cos (3p n)       for N=2  Periodic

fo = GCD (f₁, f₂)   &   T = LCM (T₁, T₂) ------- For Analog/digital signal

[Complex exponential and sinusoidal sequences are not necessarily periodic in ‘n’ with period ( Wo 2π/ow
) and depending on Wo, may not be periodic at all]

N = fundamental period of a periodic sinusoidal.

3. The highest rate of oscillations in a discrete time sinusoid is obtained when w = π or -π

Discrete-time sinusoidal signals with frequencies that are separated by an integral multiple of 2p are Identical.

4. Fs/2 ≤ F ≤ Fs/2

-πFs ≤2p F ≤ π Fs

-π ≤ Ω Ts ≤π

Therefore   - π ≤ w≤ π

5. Increasing the frequency of a discrete- time sinusoid does not necessarily decrease the period of the signal.

x₁(n) = Cos(πn/4)    N=8

x₂(n) = Cos (3πn) N=16 3/8 > 1/4

2 π f = 3π /8

⇒ f=3/16

• If analog signal frequency = F = 1/Ts samples/Sec = Hz then digital frequency f = 1

W =Ω T_s

2 π f = 2π F T_s =>    f =1

2π F =π/4;   2π f =π /4

F = 1/8 ;T=8;               f = 1/8       N=8

7. Discrete-time sinusoids are always periodic in frequency.

Q. The signal x (t) = 2 Cos (40π t) + Sin (60π t) is sampled at 75Hz. What is the common period of the sampled signal x (n), and how many full periods of x (t) does it take to obtain one period of x(n)?

F₁ = 20Hz               F₂ = 30Hz

f₁ =20/75 =4/15 =K1/N1         f₂ = 30/75 =2/5 =K2/N2

The common period is thus N=LCM (N₁, N2) = LCM (15, 5) = 15

The fundamental frequency F_o of x (t) is GCD (20, 30) = 10Hz

And fundamental period T =1/F0=0.1s

Since N=15

1sample ----------1/75 sec

15 sample ----------- ?         =>15/75 =0.2S

∴So it takes two full periods of x (t) to obtain one period of x (n) or GCD (K1, K2) = GCD (4, 2) = 2

Frequency Domain Representation of discrete-time signals and systems

For LTI systems we know that a representation of the input sequence as a weighted

sum of delayed impulses leads to a representation of the output as a weighted sum of

delayed responses.

Let x (n) = e^jwn

^{y (n) = h (n) * x (n)}

Let H  is the frequency domain representation of the system.

∴y (n) = H (e^jw) e^jwn               e^{jwn = eigen function of the system.}

H (e^jw) = eigen value 

Q. Find the frequency response of 1^st order system y (n) = x (n) + a y (n-1)

(a<1)

Let x (n) = e^jwn

y_p (n) = C e^jwn

C e^jwn = e^jwn + a C e^jw (n-1) 

C e^jwn [1-ae^-jw] = e^jwn

Therefore   H (e^jw) = 

Q. Frequency response of 2nd order system y(n) = x(n) -1/2 y(n-2)

x (n) = e ^jwn