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EECS 105 Fall 2003, Lecture 2 Lecture 2: Transfer Functions Prof. Niknejad Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Review of LTI Systems Since most periodic (non-periodic) signals can be decomposed into a summation (integration) of sinusoids via Fourier Series (Transform), the response of a LTI system to virtually any input is characterized by the frequency response of the system: Phase Shift Any linear circuit With L,C,R,M and dep. sources Department of EECS Amp Scale University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad “Proof” for Linear Systems For an arbitrary linear circuit (L,C,R,M, and dependent sources), decompose it into linear suboperators, like multiplication by constants, time derivatives, or integrals: d d2 y L( x) ax b1 x b2 2 x c1 x c2 x c3 x dt dt For a complex exponential input x this simplifies: 2 d d y L(e jt ) ae jt b1 e jt b2 2 e jt c1 e jt c2 e jt dt dt jt jt e e y ae jt b1 je jt b2 ( j ) 2 e jt c1 c2 2 j ( j ) c1 c2 jt 2 y Hx e a b1 j b2 ( j ) 2 j ( j ) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad “Proof” (cont.) Notice that the output is also a complex exp times a complex number: y Hx e jt c1 c2 2 a b1 j b2 ( j ) 2 j ( j ) The amplitude of the output is the magnitude of the complex number and the phase of the output is the phase of the complex number y Hx e jt c1 c2 2 a b1 j b2 ( j ) 2 j ( j ) y e jt H ( ) e j H ( ) Re[ y ] H ( ) cos(t H ( )) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Complex Transfer Function Excite a system with an input voltage (current) x Define the output voltage y (current) to be any node voltage (branch current) For a complex exponential input, the “transfer function” from input to output: y c1 c2 2 H a b1 j b2 ( j ) 2 x j ( j ) We can write this in canonical form as a rational function: n1 n2 j n3 ( j ) 2 H ( ) d1 d 2 j d 3 ( j ) 2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Impede the Currents ! Suppose that the “input” is defined as the voltage of a terminal pair (port) and the “output” is defined as the current into the port: + v (t ) Arbitrary LTI Circuit i (t ) v(t ) Ve jt V e j (t v ) i (t ) Ie jt I e j (t i ) – The impedance Z is defined as the ratio of the phasor voltage to phasor current (“self” transfer function) V V j (v i ) Z ( ) H ( ) e I I Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Admit the Currents! Suppose that the “input” is defined as the current of a terminal pair (port) and the “output” is defined as the voltage into the port: + v (t ) Arbitrary LTI Circuit i (t ) v(t ) Ve jt V e j (t v ) i (t ) Ie jt I e j (t i ) – The admmittance Z is defined as the ratio of the phasor current to phasor voltage (“self” transfer function) I I j (i v ) Y ( ) H ( ) e V V Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Voltage and Current Gain The voltage (current) gain is just the voltage (current) transfer function from one port to another port: + + v1 (t ) Arbitrary LTI Circuit i1 (t ) – Gv ( ) i2 (t ) v2 (t ) – V2 V2 j (2 1 ) e V1 V1 I 2 I 2 j (2 1 ) Gi ( ) e I1 I1 If G > 1, the circuit has voltage (current) gain If G < 1, the circuit has loss or attenuation Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Transimpedance/admittance Current/voltage gain are unitless quantities Sometimes we are interested in the transfer of voltage to current or vice versa + v1 (t ) + Arbitrary LTI Circuit i1 (t ) – Department of EECS i2 (t ) v2 (t ) – V2 V2 j (2 1 ) J ( ) e I1 I1 [ ] I 2 I 2 j (2 1 ) K ( ) e V1 V1 [S ] University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Power Flow The instantaneous power flow into any element is the product of the voltage and current: P(t ) i(t )v(t ) For a periodic excitation, the average power is: Pav i( )v( )d T In terms of sinusoids we have Pav I cos(t i ) V cos(t v )d T I V (cos t cos i sin t sin i ) (cos t cos v sin t sin v )d T I V d cos 2 t cos i cos v sin 2 sin i sin v c sin t cos t T I V 2 (cos i cos v sin i sin v ) Department of EECS I V 2 cos(i v ) University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Power Flow with Phasors Pav I V 2 cos(i v ) Power Factor I V cos( / 2) 0 Note that if (i v ) , then Pav 2 2 Important: Power is a non-linear function so we can’t simply take the real part of the product of the phasors: P Re[ I V ] From our previous calculation: P Department of EECS I V 2 1 1 * cos(i v ) Re[ I V ] Re[ I * V ] 2 2 University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad More Power to You! In terms of the circuit impedance we have: 2 V 1 1 V * * P Re[ I V ] Re[ V ] Re[ Z 1 ] 2 2 Z 2 V 2 2 Re[ Z Z * ] 2 V 2Z 2 2 Re[ Z * ] V 2Z 2 2 Re[ Z ] Check the result for a real impedance (resistor) Also, in terms of current: 2 I 1 1 * * P Re[ I V ] Re[ I I Z ] Re[ Z ] 2 2 2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Direct Calculation of H (no DEs) To directly calculate the transfer function (impedance, transimpedance, etc) we can generalize the circuit analysis concept from the “real” domain to the “phasor” domain With the concept of impedance (admittance), we can now directly analyze a circuit without explicitly writing down any differential equations Use KVL, KCL, mesh analysis, loop analysis, or node analysis where inductors and capacitors are treated as complex resistors Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad LPF Example: Again! Instead of setting up the DE in the time-domain, let’s do it directly in the frequency domain Treat the capacitor as an imaginary “resistance” or impedance: → time domain “real” circuit frequency domain “phasor” circuit Last lecture we calculated the impedance: ZR R Department of EECS 1 ZC jC University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad LPF … Voltage Divider Fast way to solve problem is to say that the LPF is really a voltage divider 1 V ZC 1 jC H ( ) o Vs Z C Z R R 1 1 jRC j C Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Bigger Example (no problem!) Consider a more complicated example: Z eff Vo ZC 2 H ( ) Vs Z eff Z C 2 Z eff R2 R1 || Z C1 ZC 2 H ( ) R2 R1 || Z C1 Z C 2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Does it sound better? Application of LPF: Noise Filter Listen to the following sound file (voice corrupted with noise) Since the noise has a flat frequency spectrum, if we LPF the signal we should get rid of the highfrequency components of noise The filter cutoff frequency should be above the highest frequency produced by the human voice (~ 5 kHz). A high-pass filter (HPF) has the opposite effect, it amplifies the noise and attenuates the signal. Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Building Tents: Poles and Zeros For most circuits that we’ll deal with, the transfer function can be shown to be a rational function n1 n2 j n3 ( j ) 2 H ( ) d1 d 2 j d 3 ( j ) 2 The behavior of the circuit can be extracted by finding the roots of the numerator and deonminator ( z1 j )( z2 j ) ( zi j ) H ( ) ( p1 j )( p2 j ) ( pi j ) Or another form (DC gain explicit) (1 j z1 )(1 j z 2 ) H ( ) G0 ( j ) G0 ( j ) K (1 j p 2 )(1 j p 2 ) K Department of EECS (1 j (1 j z ,i ) p ,i ) University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Poles and Zeros (cont) The roots of the numerator are called the “zeros” since at these frequencies, the transfer function is zero The roots of the denominator are called the “poles”, since at these frequencies the transfer function peaks (like a pole in a tent) Department of EECS “poles” H ( ) ( z1 j )( z 2 j ) ( p1 j )( p2 j ) University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Finding the Magnitude (quickly) The magnitude of the response can be calculated quickly by using the property of the mag operator: (1 j z1 )(1 j z 2 ) H ( ) G0 ( j ) (1 j p 2 )(1 j p 2 ) K G0 K 1 j z1 1 j z 2 1 j p 2 1 j p 2 The magnitude at DC depends on G0 and the number of poles/zeros at DC. If K > 0, gain is zero. If K < 0, DC gain is infinite. Otherwise if K=0, then gain is simply G0 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Finding the Phase (quickly) As proved in HW #1, the phase can be computed quickly with the following formula: K (1 j z1 )(1 j z 2 ) H ( ) G0 ( j ) (1 j p 2 )(1 j p 2 ) G0 ( j ) K (1 j z1 ) (1 j z 2 ) (1 j p1 ) (1 j p 2 ) No the second term is simple to calculate for positive frequencies: ( j ) K K 2 Interpret this as saying that multiplication by j is equivalent to rotation by 90 degrees Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Bode Plots Simply the log-log plot of the magnitude and phase response of a circuit (impedance, transimpedance, gain, …) Gives insight into the behavior of a circuit as a function of frequency The “log” expands the scale so that breakpoints in the transfer function are clearly delineated In EECS 140, Bode plots are used to “compensate” circuits in feedback loops Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Example: High-Pass Filter Using the voltage divider rule: L j R jL H ( ) R j L 1 j L R j H ( ) 1 j j H 1 j 0 0 H 0 1 0 1 j 1 H 1 j 2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad HPF Magnitude Bode Plot Recall that log of product is the sum of log H ( ) dB j j 1 j dB 1 j dB 1 j dB Increase by 20 dB/decade 1 j dB 0 dB Equals unity at breakpoint dB 40 dB 20 dB 20 dB 0 dB 0.1 1 10 100 -20 dB Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad HPF Bode Plot (dissection) The second term can be further dissected: 1 1 j 0 dB .1/ -20 dB 1/ 0 dB 1 j dB dB 10 / 1 1 1 20 dB -40 dB -60 dB 0 dB -20 dB/dec ~ -3dB - 3dB Department of EECS 1 University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Composite Plot Composit is simply the sum of each component: 1 j dB 1 j j dB dB 0 dB .1/ 1/ High frequency ~ 0 dB Gain 10 / Low frequency attenuation -20 dB 1 1 j dB -40 dB Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad Approximate versus Actual Plot Approximate curve accurate away from breakpoint At breakpoint there is a 3 dB error Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 2 Prof. A. Niknejad HPF Phase Plot Phase can be naturally decomposed as well: j 1 H ( ) j tan 1 1 j 1 j 2 First term is simply a constant phase of 90 degrees The second term is a classic arctan function Estimate argtan function: 1 -45 1 Actual curve -90 Department of EECS 1 University of California, Berkeley