ECE 3336 Introduction to Circuits & Electronics

1. ECE 3336 Introduction to Circuits & Electronics Note Set #11 Frequency Response Fall 2013, TUE&TH 4:00-5:30 pm Dr. Wanda Wosik

2. Frequency Dependence is Important in Electronics Frequency dependence is seen in many non-electrical 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

3. Frequency Response Will Rely onPhasor Analysis and Impedances 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 steady-state 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.

4. 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. • Depending on a circuit, some input signals may appear at the output with smaller amplitudes or not appear at all. We say that these signals are attenuated. • In addition, thephase shiftcauses additional signal distortion (as below). Frequency Response 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).

5. How to Deal with Frequency Dependence in Circuits? • If multiple frequencies, such as those from a CD player, are included in the input signal, we may have a distortion at the output since some frequencies can be lost by attenuation. 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.

6. Use Phasors to Find the Frequency Response 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

7. 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

8. Periodic Signals Represented by Fourier Series 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.

9. Fourier Analysis Any periodic signal will be represented by superposition of infinitive number of sinusoidal signals expressed by a Fourier Series Sine-cosine representation Magnitude and phase representation In Fourier Series: Fundamental frequency 0=2f0=2/T Harmonics: 20,30, 40 Equivalent coefficients

10. 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 Series Example: 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 non-sinusoidal 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

11. Other Examples of Signals Represented by a Fourier Series 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

12. More Examples of Signals Represented by a Fourier Series 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

13. Response of Linear Systems to Periodic Input Signals 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).

14. Filters • Filters are very important circuits that allow for selective attenuation of signals of a specific frequency or of specific frequency ranges. All other signals with other frequencies will pass without attenuation. • Magnitudeof the transfer function |H(j)| affects the amplitude of the output signal. • Phaseof thetransfer function |H(j)| corresponds to the phase shift between the output and input signals. A two port circuit where H(jw) Is amplitude Is phase

15. Filters • The frequency blocks marked in red in the figure below, indicate the passing frequencies. Signals with all other remaining frequencies will be stopped by the circuit so they will not appear at the output. |H(j)| |H(j)| low stopband high stopband  pass pass  A low-pass filter passes Only low frequency signals A high-pass filter passes only high frequencies

16. Other filters 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

17. Which Filters Will We Cover Here? • First order filters operating as low-pass and high-pass filters. These are made either as RC and RL filters. That means that capacitors and inductors will not be included simultaneously. • Bandpass and resonant filters. They will have (usually, but not always) all three elements included: an inductor, capacitor and resistor. • Active filters. They have very similar operation but they also can amplify the amplitude of the signals that pass through the circuit. They are built using operational amplifiers. This will be done later in the course (Notes #13&14).

18. Low-pass filter Output Slow will pass Fast will be stopped

19. Low-pass Filter 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/jC) Therefore the transfer function is:

20. Low-pass filter Our transfer function will be now expressed using the capacitance and resistance. AND We will find its magnitude and phase. Phase Magnitude • We will select a new parameter called: cutoff frequency=1/RC. • This will be a reference pointfrom which we will monitor amplitude and phase changes in out output signal.

21. Frequency Response for the Low-pass 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

22. High-pass Filters Output Fast will pass Slow will be stopped

23. High-pass Filter 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 low-pass 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/jC) Therefore the transfer function is:

24. We will find the magnitude and phase of H(j). Frequency Behavior of a High-pass Filter phase magnitude • As for the low-pass filter • We will again select a new parameter called: cutoff frequency=1/RC. • This will be a reference pointfrom which we will monitor amplitude and phase changes in our output signal.

25. We will first rewrite the expression for the transfer function by introducing 0=1/RC Magnitude and Phase of the Transfer Function Calculate 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 half-power 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

26. Bode Plots 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

27. Towards Derivation of Bode Plots Back to a low-pass 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 phaseH(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:

28. Derivation of Bode Plots 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

29. 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(j-tan-1(1/10)=0° For >> 0 =0 H(j-tan-1(1)=-45° =100 |H(j)|dB=-20log(10)=-20 dB =100 H(j j-tan-1(10)=-90° |H(j)|dB=-20log(100)=-40 dB =1000 0° =10000 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 -

30. 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(j90°-tan-1(1/10)=0° =0/10 =0/10|H(j)|dB=-20log(10-1)=-20 dB H(j90°-tan-1(1)=45° =0 H(j j90°-tan-1(10)=0° |H(j)|dB=-20log(10-2)=-40 dB =100 =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

31. 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 high-pass 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

32. Differentiator • Frequency of the input signal changes • from low • through medium • to high http://www.electronics-tutorials.ws

33. 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 high-pass 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

34. Integrator • Frequency of the input signal changes • from low • through medium • to high http://www.electronics-tutorials.ws