Objectives Boise State University Department of Electrical and Computer Engineering ECE L Circuit Analysis and Design Lab Experiment #0: Frequency esponse Measurements The objectives of this laboratory experiment are: To obtain the frequency response (magnitude and phase) of a two-port network. To identify the transfer function of a minimum-phase network from the poles and zeros of the corresponding Bode magnitude plot. Theory X(s) H(s) Y(s) ~ X H(j ω) ~ Y (a) (b) Figure : (a) Transfer Function and (b) its Frequency esponse Consider the block diagram representation of a transfer function in Figure (a). ecall that the transfer function of a linear circuit is defined as the ratio of the output response Y (s) to an input signal X(s) with all initially conditions set equal to zero, i.e., with all initial inductor currents and all initial capacitor voltages set equal to zero. Mathematically, H(s) = Y (s) X(s) The input X(s) can be a voltage source or a current source. The output Y (s) is usually a voltage or a current output. The transfer function H(s) has different names depending on the choice of input and output variables as shown in Table. Table : Descriptive Names of Transfer Functions Input X(s) Output Y (s) Transfer Function H(s) = Y (s)/x(s) Voltage Source Voltage Output Voltage Gain Voltage Source Current Output Admittance Current Source Voltage Output Impedance Current Source Current Output Current Gain Usually, the transfer function of a linear circuit is derived as a rational fraction of two polynomials in the Laplace variable s. The numerator and denominator of H(s) can be factored out to yield the zeros (roots of the numerator of H(s)) and poles (roots of the denominator of H(s)) as follows: H(s) = N(s) D(s) = K(s z ) (s z m ) (s p ) (s p n ) () ()
Under the form of Equation (), the transfer function H(s) has m zeros equal to z,..., z m and n poles equal to p,..., p n. If a sinusoidal signal x(t) = X m cos(ωt θ i ) is applied to the input of a linear circuit, then its output response y(t) = Y m cos(ωt θ o ) will be sinusoidal after all the circuit transients have died out. As the frequency is varied from zero to infinity, the ratio of the output phasor Ȳ = Y m θ o to the input phasor X = X m θ i is called the frequency response of the linear circuit H(jω) = H(ω) θ(ω) = Ȳ X = Y m θ o X m θ i = Y m X m (θ o θ i ) (3) The magnitude H(ω) of the frequency response can be obtained from the ratio of the peak-to-peak output voltage to the peak-to-peak input voltage as follows: H(ω) = Y m X m = Y m X m = Y pp X pp (4) The phase θ(ω) of the frequency response can be measured from the time delay of the output sinusoid with respect to the input sinusoid as follows: θ(ω) = θ o θ i = 360 o t T The relationship between a transfer function and its frequency response is shown in Figure (b). It can be shown that the transfer function H(jω) = H(ω) θ(ω) can be obtained from the transfer function H(s) by replacing the Laplace variable s by jω, that is, H(jω) = H(ω) θ(ω) = [H(s)] s = jω (6) and converting the resulting complex expression into polar form (magnitude and angle). The frequency response can be visualized with the plots of the magnitude and phase of H(jω) as a function of frequency ω (rad/s) or f (Hz). If the frequency response of a circuit is known, it will allow the prediction of the output response to any periodic input, whether sinusoidal or nonsinusoidal. For example, a periodic input signal such a square or triangular waveform can be decomposed as a sum of sinusoids called its Fourier series representation. The output response of each sinusoidal component can be found from the definition of the frequency response at a particular frequency ω as Ỹ = Y m θ o = H(jω ) X = H(ω )X m (θ i θ(ω )) (7) yielding the output sinusoidal waveform y(t) = Y m cos(ω t θ o ) = H(ω )X m cos(ω t θ i θ(ω )) (8) Equations (7) or (8) indicate how the particular sinusoidal component at the frequency ω = ω will be effected by the frequency response of the circuit. It is clear that the magnitude Y m of the output response will be equal to the magnitude X m of the input signal multiplied by the gain H(ω ) of the frequency response at ω = ω and the phase θ o of the output response will be shifted from the phase θ i of the input signal by the phase θ(ω ) of the frequency response at ω = ω. Another representation of the frequency response of a circuit is in terms of the Bode magnitude and phase plots. The Bode magnitude in decibels (db) is defined by H (db) = 0 log 0 H(ω) (9) (5)
where log 0 is the logarithm of base 0. This magnitude in db is plotted as a function of the logarithm of base 0 of the frequency ω (rad/s) or frequency f (Hz) on semilog paper. The phase θ(ω) of the frequency response is also plotted as a function of the logarithm of base 0 of the frequency ω (rad/s) or frequency f (Hz) on semilog paper. A knowledge of the simple poles and zeros of the transfer function H(s) allows a quick manual sketch of these plots which are referred to as the Bode straightline approximations. sc sc V (s) i V o (s) (a) H (db) 0 7 0 0 db/dec 0 db/dec 0.f f f 0f f (Hz) (b) 90 θ (deg) 45 0 45 45 deg/dec 0.f f 90 deg/dec f 0f f (Hz) 45 deg/dec 90 (c) Figure : (a) Bandpass Filter and (b) its Bode Magnitude and (c) Bode Phase Plots In this experiment, the transfer function of the op-amp circuit shown in Figure (a) will be identified from frequency response measurements. The circuit in Figure (a) is a bandpass filter. The transfer 3
function of this circuit can be shown to equal H(s) = V o(s) V i (s) = sc ( sc )( sc ) = Ks ( s/ω )( s/ω ) where K = C, ω = /C, and ω = /C. The magnitude and phase of the frequency response H(jω) are found as K(jω) H(jω) = H(ω) θ(ω) = ( jω/ω )( jω/ω ) Kω = (ω/ω ) ( (ω/ω ) 90 o tan ω tan ω () ω ω The straightline Bode magnitude plot is shown in a solid line in Figure (b). The low-frequency asymptote starts with a slope of 0 db/dec due to the pure differentiator in the numerator. The gain then flattens after encountering the first pole at ω and then drops with a slope of -0 db/dec after encountering the second pole at ω. The low-frequency asymptote crosses the midband gain at the corner frequency f while the high-frequency asymptote crosses the midband gain at f. The exact Bode magnitude plot is shown in a dashed line and has a maximum error of -3 db at the corner frequencies f and f. In other words, the actual amplifier gain has a maximum error of about 3 db at these corner frequencies which are also known at the 3-dB frequencies. The gain in actual units (V/V) is / or 0.707 of the midband gain at these corner frequencies. The straightline Bode phase plot is shown in a solid line in Figure (c). The phase plot starts flat at 90 o degrees one decade below the corner frequency f and then drops off at a rate of -45 deg/dec for one decade. Between the frequencies f and f = 0f, the slope is -90 deg/dec. Finally, the straight-line slope is -45 Deg/dec before leveling off at 90 o. The exact Bode phase plot is shown in a dashed line and has a maximum error of about 6 degrees at f /0 and at 0f. 3 Equipment Agilent DSO504A Digital Storage Oscilloscope Agilent 330A Function/Arbitrary Waveform Generator Fluke 5 True MS Multimeter Assorted resistors, capacitors, and op-amps 4 Procedure. Measure the following components using your benchtop multimeter. Table : Parameter Measurements (0) C C Nominal 00n 6k n 60k Measured. Wire the op-amp circuit in Figure (a) on your protoboard. Set your function generator on HIGH Z and apply a -V peak-to-peak sine waveform with variable frequency and no offset to the input of the C network. Set up the infinium scope to record the following quantities: 4
Table : Frequency esponse Measurements f (Hz) V i,pp (V) V o,pp (V) t (µs) θ = 360 o t/t (deg) 0 0 40 60 80 00 00 400 600 800 k k 4k 6k 8k 0k 5 Data Analysis and Interpretation. Derive the transfer function shown in Eq. (0).. Plot the exact Bode magnitude and phase plots for the op-amp filter of Figure (a) on two separate graphs. Compare the shape of these graphs with those shown in dashed lines in Figures (b) and (c). 3. On the magnitude plot, manually draw the midband gain and read off its value from the exact plot. 4. Draw the low-frequency and high-frequency asymptotes on the Bode magnitude plot. Verify that they intersect the midband gain at the 3-dB corner frequencies f (Hz) and f (Hz) which are 3 db below the midband gain on the exact magnitude plot. 5. From the knowledge of f (Hz) and f (Hz), add manually the Bode straight-line phase approximations. ead off the exact values of the phase at the corner frequencies f (Hz) and f (Hz). 6. Using the measured values for and, deduce the unknown capacitor values C (µf) and C (nf) from the corner frequencies f (Hz) and f (Hz). 5
Boise State University Department of Electrical and Computer Engineering ECE L Circuit Analysis and Design Lab Experiment #0: Frequency esponse Measurements Date: Data Sheet ecorded by: Equipment Agilent DSO504A Digital Storage Oscilloscope Agilent 330A Function/Arbitrary Waveform Generator Fluke 5 True MS Multimeter BSU Tag Number or Serial Number Table : Parameter Measurements C C Nominal 00n 6k n 60k Measured Table : Frequency esponse Measurements f (Hz) V i,pp (V) V o,pp (V) t = t t (µs) θ = θ θ (deg) 0 0 40 60 80 00 00 400 600 800 k k 4k 6k 8k 0k