ECE 3336 Introduction to Circuits & Electronics. Note Set #11 Frequency Response . Fall 2013, TUE&TH 4 : 0 05: 3 0 pm Dr. Wanda Wosik. Frequency Dependence is Important in Electronics.
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Note Set #11
Frequency Response
Fall 2013,
TUE&TH 4:005:30 pm
Dr. Wanda Wosik
Frequency dependence is seen in many nonelectrical phenomena (resonant vibrations, mechanical oscillations, damping, rotations and many, many others).
Frequency dependence is also “frequently” present and very important in electronic circuits.
Electrical signals of various frequencies used for various applications are being processed by circuits designed specifically to operate at these frequencies.
Examples include dependences between voice frequency and tone , frequency and color of light
We will study behavior of circuits when signal frequency changes.
We will use phasors and impedances to analyze acsignals that are delivered to the circuits and propagate there.
Circuits will perform various functions, and finally the signals will appear (or not i.e. will disappear) at the output.
We would like to monitor these ac signals in circuits that operate in the steadystate conditions.
We have to develop a tool that would let us predict the behavior of the output signalfor any input signal that is supplied to the input.
This means that, at the beginning, we will be rather analyzing circuits not the specific signals.
Clear? Probably NOT yet.
Signals of various frequencies and amplitudes applied to the circuit input may appear at the output terminals with different amplitudes and with their phases shifted compared with the input signal.
Attenuation and distortion are the consequences of capacitive and inductive behavior of various circuit elements.
They can be just simple capacitors and/or inductors in the circuit but more frequently there are parasitics that result in such effects.
(Fig. 6.15 from Rizzoni’s book).
Load
Source
So, it is wise to determine a Frequency Response or a Transfer Function Hv(j)that will work for a given circuit. This transfer function will affect any ac steady state signal applied to the input.
To predict the frequency behavior of a circuit, we define the transfer function that relates the output (load) voltage to the input voltage in the frequency domain.
Th. Eq.
Now, to find VL we will first find Thevenin equivalent
Transfer Function Characterizes
a Circuit not a Signal
The transfer function HV(j) is a complex number that has a magnitude andphase
Once we derive HV(j) for a circuit
(NOT FOR A SIGNAL!)
we will find an output signal VL(j) for any input signal Vs(j) .
The output voltage (load) VL(j)is calculated from the source voltage Vs(j) multiplied by the transfer function using the complex numbers rules.
where
Is amplitude
Is phase
The goal is to find the Transfer Function for periodic signals.
A periodic signal is shown below.
n=1, 2, 3, ….; T=period
We will use Fourier theorem, which allows for representation of periodic signals as a superposition of various sinusoidal components of various frequency and different amplitudes.
Any periodic signal will be represented by superposition of infinitive number of sinusoidal signals expressed by a Fourier Series
Sinecosine representation
Magnitude and phase representation
In Fourier Series:
Fundamental frequency 0=2f0=2/T
Harmonics:
20,30, 40
Equivalent coefficients
For any signal, the coefficients a0, an and bn have to be calculated in order to use Fourier series to represent this signal.
Square Wave Expanded in Fourier SeriesExample: For a square wave, which is odd function an=0 while bn≠0.
So the signal will be represented by a sum of sine waves:
Each consecutive harmonic has a decreasing amplitude and increasing frequency.
Increasing number of harmonics allows for improved representation of the nonsinusoidal signals. Infinitive number of harmonics will give a perfect match.
You may want to see More Examples of Signals Represented by a Fourier Series http://www.stat.ucla.edu/~dinov/courses_students.dir/04/Spring/Stat233.dir/Stat233_notes.dir/JavaApplet.html to play with harmonics
Sawtooth function
From the calculated coefficients an=0 and
we have v(t)
Notice how the amplitude of harmonicsdecreases
We can calculate the amplitudes cn and the phases n
Coefficients
Pulse train signal
Amplitude of harmonics
Matching will improve with the number of harmonics included in the Fourier Series
Phase of harmonics
You may want to see More Examples of Signals Represented by a Fourier Series http://www.stat.ucla.edu/~dinov/courses_students.dir/04/Spring/Stat233.dir/Stat233_notes.dir/JavaApplet.html to play with harmonics
Periodicsignals applied to the input of a circuit, which has capacitors or inductors (real or parasitic) will be modified by this circuit.
The output signal will have modified amplitude and phase compared to the original signal at the input.
We will use the Fourier series to represent such signals.
If we take a sinusoidal signal at the input of a linear system, represented by a finite number of components
we can trace the Qout(j) output signal if we know both the input signal Qin(j) and the transfer function H(j).
A two port circuit
where
H(jw)
Is amplitude
Is phase
H(j)
H(j)
low
stopband
high
stopband
pass
pass
A lowpass filter passes
Only low frequency signals
A highpass filter passes only high frequencies
A very important family of filters include resonant circuits, which have quite significant applications of electronics.
They select specific frequencies to be either passed or frequencies to be attenuated.
They usually require all three elements: capacitors, inductors, and resistors. We will call them second order filters.
H(j)
H(j)
band
band
Bands can be broad
rejection
pass
H(j)
H(j)
Very narrow bands
passing
rejection
Resonance filters
To monitor the frequency dependence of this filter we need to derive H(j)
To find H(j) we have to find V0(j) as a a function of the input signal Vi(j).
So we will use a voltage divider
Now we can estimate the magnitude of H(j) at very low and very high frequency:
For =0 (DC conditions) H(j)=1
Here the capacitor acts as an open circuit
For ∞ H(jw) 0
The capacitor acts as a short (Z=1/jC)
Therefore the transfer function is:
Our transfer function will be now expressed using the capacitance and resistance.
AND
We will find its magnitude and phase.
Phase
Magnitude
Frequency Response for the Lowpass Filter
We will now find the magnitude and phase of H(j) usingthe expression for the breakpoint frequency=1/RC.
.
We will calculate the
Magnitude of the transfer function
0.707
At 0
0
We will then calculate the
Phase of the transfer function
45°
At ~0 Phase=0°
At 0 Phase=45°
At 0 Phase=90°
0
To monitor the frequency dependence of this filter we need to derive the Transfer Function H(j). We will use the same approach as we used for the lowpass filter.
To find H(j) we have to find V0(j) as a function of Vi(j).
So again we will use a voltage divider
Now we can estimate the magnitude of H(j) at very low and very high frequency:
For =0 (DC conditions) H(j)=0
Here the capacitor acts as an open circuit
For ∞ H(jw) 1
The capacitor acts as a short (Z=1/jC)
Therefore the transfer function is:
We will find the magnitude and phase of H(j).
Frequency Behavior of a Highpass Filterphase
magnitude
We will first rewrite the expression for the transfer function by introducing 0=1/RC
Magnitude and Phase of the Transfer FunctionCalculate the
Phaseof the transfer function
Calculate the
Magnitudeof the transfer function which at 0 is:
At ~0 Phase=90°
At 0 Phase=45°
At 0 Phase=0°
That's why 0 is also called halfpower frequency. Here Vo(j)~H(j) and the power delivered by V0 is divided by
Notice: The phase shift coincides with the change in the magnitude. No change in the magnitude no change in the phase.
45°
0.707
Amplitude
phase
Bode Plots will be used to simplify graphical representation of the magnitude and phase of the transfer functions. We will use them instead of plotting directly the results from calculation of the H(j) and H(j).
Bode Plots will give us an approximation of the transfer function in the broad range of frequency changes. Accuracy will be acceptable even at breakpoint frequencies.
These are straight line approximations both for the magnitude and phase of the Transfer Function. That allows us to easily predict its frequency dependence.
Magnitude
note the linear scale
Phase
Back to a lowpass filter composed only of one capacitor and one resistor.
Earlier we have derived its transfer function:
We also calculated and plotted its magnitude H(j) and phaseH(j).
0.707
magnitude
45°
phase
Now we will use new approximations for the magnitude and the phase instead of calculating step by step the transfer characteristics.
To do that we will now use decibels (dB) defined as:
These are approximations for the magnitude and the phase – accurate ones!
Start with expressing the ratio of magnitudes of two complex numbers
using decibels (dB), which are defined as:
Since the transfer function H(j) is a ratio of the two voltages and it is a complex number, we will calculate the magnitude of this number H(j)
using decibels. We will obtain H(j)dB
Plotting Bode Plots of the Transfer Function (low pass)
We will plot the MAGNITUDE at selected multiples of 0.
Then we will plot the PHASE
For <<0 H(j)dB=20log10(1)=0 dB
At =0 H(j)dB=20log10√(1+1)=3 dB
=0/10
H(jtan1(1/10)=0°
For >> 0
=0
H(jtan1(1)=45°
=100 H(j)dB=20log(10)=20 dB
=100
H(j jtan1(10)=90°
H(j)dB=20log(100)=40 dB
=1000
0°
=10000
etc.


The influence of 0 is seen in H(j) for all subsequent frequencies.
The 0 affects the phase only locally: within two decades only

20dB/dec
45°
45°/dec



0
0

90°

also known as 3dB frequency

Plotting Bode Plots of the Transfer Function (high pass)
Plot the MAGNITUDE at selected fractions of 0.
Plot the PHASE
At =0 H(j)dB=20log10√(1+1)=3 dB
H(j90°tan1(1/10)=0°
=0/10
=0/10H(j)dB=20log(101)=20 dB
H(j90°tan1(1)=45°
=0
H(j j90°tan1(10)=0°
H(j)dB=20log(102)=40 dB
=100
=0/100
>>0
H(j)dB=0
The influence of 0 is seen in H(j) for all frequencies <<0.
The 0 affects the phase only locally: within two decades only
+20dB/dec
0
aka 3dB frequency
How Charging of a Capacitor Relates to the Frequency Response
Example: High pass filter
We start with the current, which is the same through the capacitor and resistor
Which for low frequencies i.e. when <<otherefore vi(t)>>vo(t), becomes
So the highpass filter acts then as a differentiator.
RC
Short time constant RC gives distortion; a square wave with a very small amplitude appears at the output.
Increasing RC will make both the signals increasingly the same
Increasing time constant RC
or equivalently: decreasing
=1/RC breakpoint frequency of the circuit, makes the output signal lessdistorted compared to the input signal of a specific frequency. Exponential decay.
RC
http://www.electronicstutorials.ws
How Charging of a Capacitor Relates to the Frequency Response
Example: Low pass filter
We start again with the current, which is the same through the resistor and capacitor
Which for high frequencies i.e. when >>o therefore vi(t)>>vo(t), becomes
Long RCtime constant gives distortion i.e. a triangular wave with a very small amplitude appears at the output
But decreasing RC will make both the input and output signals increasingly the same.
So the highpass filter acts then as a integrator.
Decreasing time constant RC
or equivalently: increasing
=1/RC breakpoint frequency of the circuit, makes the output signal less distorted compared to the input signal of a specific frequency.
RC
RC
http://www.electronicstutorials.ws