Frequency response. Pavel Máša  XE31EO2. XE31EO2 Lecture11. Pavel Máša  XE31EO2  Frequency response


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1 Frequency response XE3EO2 Lecture Pavel Máša  Frequency response
2 INTRODUCTION Frequency response describe frequency dependence of output to input voltage magnitude ratio and its phase shift as a function of frequency It can describe e.g.: Which frequencies pass filter Crossover in loudspeaker system Tuner in radio / TV, ADSL splitter Anti aliasing filter in CD player Which frequency range can be played with audio amplifier Affects maximum refresh frequency and screen resolution of analog monitor (it is related to frequency response of input amplifier)  Frequency response
3 TRANSFER FUNCTION AND ITS GRAPHICAL REPRESENTATION Consider linear 2 port Frequency dependence of output voltage may be qualified by transfer function Transfer function is a function of variable ω; graphical representation may be either:. In complex plane as Nyquist plot Im 0,5 ω 0,5 0 0,5 Re 0,5 ω=0 ω Following graphs represents frequency response of integrating RC circuit It s a curve in a complex plane, where to distinct points are assigned distinct values of frequency ω it is a path allocated by phasor endpoint, when frequency varies from zero to infinity The distance of selected point (related to distinct frequency ω) from origin defines the modulus of transfer function The angle between real axis and phasor is phase of the transfer function Nyquist plot of passive circuits with the exception of resonant circuit is situated inside unit circle It is important for examination of stability of circuits with feedback (Nyquist stability criterion) in practice it is simpler measure frequency response of the circuit than search for poles of (unknown) transfer function  Frequency response
4 2. Or separated into two parts as modulus and phase frequency response [db] [rad] Modulus frequency response is drawn as Both axes of modulus frequency response are logarithmic The unit is decibel [db] Phase frequency response is drawn as Frequency axis is logarithmic, phase axis is linear The unit is radian [rad]  Frequency response
5 BODE PLOT In the 930s Hendrik Wade Bode proposed simple, but accurate method for graphing modulus and phase responses By this method is possible draw accurate responses without use of computer Frequency response has information about time constants (in transients), quality factor of resonant circuit etc. Transfer function is generally P M k=0 P (p) = b kp k P N k=0 a kp = b 0 + b p + b 2 p b M p M k a 0 + a p + a 2 p a N p N Q M k= = K (p z k) Q N k (p p k) = b M (p z )(p z 2 ) (p z M ) a N (p p )(p p 2 ) (p p N ) z k p k roots of the polynomial in nominator zeros Roots of the polynomial in denominator poles here are hidden time constants of the transient By partial fraction decomposition of this transfer function we may find transient response The graph of this transfer function is spatial surface above p plane But we are interesting in the cut through the p plane, when ¾ =0 p = ¾ + j =0+j = j Then we get to sinusoidal steady state Variable is not frequencyω, but complex frequency jω  Frequency response
6 pole (tends to ) P(j) = P M k=0 b k(j) k P N k=0 a k(j) k = frequency response yellow cut is the modulus frequency response b 0 + b j + b 2 (j) b M (j) M a 0 + a (j)+a 2 (j) a N (j) N Q M k= = K (j z k) Q N k (j p k) = b M (j z )(j z 2 ) (j z M ) a N (j p )(j p 2 ) (j p N )  Frequency response
7 The principle of the Bode plot are properties of logarithm, namely: Logarithm of the product is the sum of logarithms Logarithm of the quotient is the difference of logarithms log = 0 Considering last property is necessary normalize brackets in factorization: μ j z k = z k j + z k Then, if, log ³ j + 0 z k z k μ À, log ³ j + log z k z k z k Q M P 0 (j) = K 0 k= (j z k +) Q N k (j p k +) = b M( z )( z 2 ) ( z M ) a N ( p )( p 2 ) ( p N ) (j k= Graphically line with the slope 20 db / decade z M +) (j p +)(j p 2 +) (j p N +) z +)(j z 2 +) (j Modulus response: MX F db () =20log( P 0 (j) ) =20log(K 0 NX ))+ 20 log μ j + 20 log μ j + z k p k Phase response: '() =arg(p 0 (j)) = MX arg k= μ j + z k NX arg k= μ j + p k k=  Frequency response
8 Now we will investigate frequency response of RLC circuit the same, on which we studied 2 nd order transients P (p) =. R = 4 kω pc pl + R + pc P (p) = = L = H C = μf R = 4 kω, 2 kω a kω p 2 LC + prc + = LC p 2 + p R L + p p + P(j) = p ;2 = 2000 p = : = 3732: = 267:9 P 0 (j) = 3732: 267:9 (j 3732: +)(j LC (j) (j)+ 267:9 +) = (j (j) ;2 = 3732: = 267:9 3732: +)(j 267:9 +)  Frequency response
9 2. R = 2 kω P (p) = p p + P(j) = (j) (j)+ p ;2 = 000 p = 000 (j) ;2 = 000 P 0 (j) = 3. R = kω P (p) = (j = 000 +)2 (j 000 +)2 p p + P(j) = p ;2 = 500 p = j (j) (j)+ P 0 (j)= (j 000 )2 + j = (j 000 )2 + j (j) ;2 = j r Q  Frequency response
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