Digital Signal Processing Reference
In-Depth Information
Thus, he determined that the amplitude ratio is
ƒV
2
(v
x
) ƒ
ƒV
1
(v
x
) ƒ
ƒH(v
x
) ƒ =
= A
2
.
By measuring and recording the time lag between zero-crossing points on the output sinusoid
relative to the input, the student was able to calculate the phase difference, using the equation
t
1
-
t
2
T
x
f
2
=
* 360°.
As shown in Figure 5.26(c),
t
1
and
t
2
are the times, of the zero crossings of the input and out-
put waveforms, respectively, and
T
x
is the period of the waveforms at the frequency currently
being used.
To confirm that the data were correct, the student reviewed his circuit analysis notes
and then derived the transfer function of the network, applying
sinusoidal-steady-state
circuit
analysis techniques:
j(R/L)v
R
R + jvL + 1/jvC
=
H(v) =
-v
2
+ 1/LC + j(R/L)v
j10
4
v
-v
2
+ 10
8
+ j10
4
v
=
1
=
- 10
8
)/10
4
v
.
1 + j(v
2
The student wrote the transfer function in polar form as
H(v) = ƒH(v) ƒ e
jf(v)
,
where
1
ƒ
H(v)
ƒ
=
[1 + [(v
2
- 10
8
)/10
4
v]
2
]
1/2
and
f(v) =-tan
-1
[(v
2
- 10
8
)/10
4
v].
The student then plotted the magnitude and phase versus frequency, as shown in Figure 5.27.
The experimental data from the physical system compared reasonably well with the
calculated data, but did not coincide exactly with the plots of Figure 5.27. This is to be ex-
pected, since the mathematical model used to compute the data can never describe the phys-
ical system perfectly.
■
A widely used technique for displaying the frequency response of systems is the
Bode plot. Bode plots most commonly consist of one semi-logarithmic plot of the mag-
nitude frequency response (in decibels) and a separate semi-logarithmic plot of the
phase frequency response (in degrees), both plotted versus frequency on
log
10
scale.
