Laboraory # Chap 3 Objecives Linear Sysem Response: general case Undersand he difference and he relaionship beween a sep and impulse response. Deermine he limis of validiy of an approximaed impulse response. The idealized impulse funcion has some of he following properies : δ( ) lim a 0 R-C firs order low-pass filer: H( jω) jωτ u( ) u( a) a δ( ) d d u( ) τ ==> H( s) ==> sτ s τ h( ).. τ e τ u( ) Funcion Laplace Transform Fourier Transform f() F(s) F(jω) d f()/d s F(s) f(0 + ) jω F(jω) f( τ)dτ F(s) / s F(jω) / jω δ() /s - e -a /(s+a) /(jω+a) Sep Response: Impulse Response: Y(s) = H(s) / s ==> y() = h(τ) dτ h() = d y() /d ==> The linear sysem impulse response is he derivaive of is sep response ADSP/Lab 3. e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Procedure ) Build he following R-C circui: Wih: R = Ω, C = nf ==> τ = R C = s 2) u() is a square-wave wih u low = 0V and u high = V, 50% duy cycle. Choose an appropriae frequency in order o make sure ha he period is a leas equal o 0 τ. Verify ha u2() is equal o he inegral of his circui impulse response! i.e: u2() = e -/τ (rising edge a = 0) u2() = e -(-o)/τ (falling edge a = o) 3) Wha happens if he square-wave frequency is no chosen correcly? 4) u() is a recangle pulse wih u low = 0V and u high = V. ON-ime: on = τ / 0 OFF-ime: off = 0 τ Verify ha u2() is approximaely equal o he impulse response of his circui Remember he scaling facor!!! 5) Wha happens if on and off are no chosen correcly? Consider several siuaions. ADSP/Lab 3. 2 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Objecives Laboraory #2 Chap 3 Linear Sysem Response: general case Apply he Laplace Transform heory in a pracical example. Undersand he relaionship beween a filer bandwidh and is rise and/or fall ime (ransien mode) From Laplace Transform heory: u2() = L - [L[u()] L[h()]] = L - [U(s) H(s)] h(): second order band-pass filer (Rp ), u(): uni sep: => U(s) = /s SL.. SC. SL. SC H( s). CR.. s SL.. SC R. s 2 s => Y( s). CR. LC. s H( s) CR.. s s 2 s CR. LC. SL. SC. ωo From Laplace Transform able: k. => ke. a.. sin( ωo. ) ke. τ. sin( ωo. ) s 2 2 a ωo 2 Afer some algebra: k Example: 0. CR. 2 L 4 a 2C.. R R = 47 k, C = 27 nf, L = 0.5 H τ 2C.. R ωo LC. 4C. 2. R 2 u2 ( ) 0.05 0 0.05 0. 0 0.002 0.004 0.006 0.008 0.0 k = 0.092 τ = 2.53 ms fo =.37 khz ADSP/Lab 3. 3 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Procedure ) Build he following second order R-L-C band-pass filer: R = kω L = H C = nf ==> fo = khz, Bw /(2 π Req C) = Hz (wih Q = fo/bw > 5) 2) u() is a square-wave wih u low = 0V and u high = V, 50% duy cycle. Choose an appropriae frequency in order o make sure ha he period is a leas equal o 5/Bw. Measure k, τ and fo and compare your resuls wih he heoreical values. 3) Wha happens if he square-wave frequency is no chosen correcly? 4) Wih he same signal generaor parameers, divide R by 3. Wha happens? 5) Wih he same signal generaor parameers, divide R by 0. Wha happens? 6) Impulse response measuremen: From heory, we know ha: d h() y sep d () τ ke.. ωo. cos( ωo. ). sin( ωo. ) τ if ωo> 0. => h approx () k. ωo. τ e. cos( ωo. ) τ u() is a recangle pulse wih u low = 0V and u high = V. ON-ime: on = / (20 fo) OFF-ime: off = 5 Bw Verify ha u2() is approximaely equal o he impulse response of his circui Remember he scaling facor!!! 7) Wha happens if on and off are no chosen correcly? Consider several siuaions. 8) Apply a FSK signal (VCF inpu of your funcion generaor driven by a squarewave of appropriae ampliude and frequency!) o your band-pass filer. Deermine he fases daa rae allowing daa recovery. Conclusion? Choose: freq high.5 freq low, freq high = fo ADSP/Lab 3. 4 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Objecives Laboraory #3 Chap 3 Linear Sysem Response: general case Undersand muliplicaion in he ime domain and he corresponding convoluion in he frequency domain. Verify ha he produc of wo periodic signals generaes sums and differences of frequencies conained in he periodic signals. y() = x() x2() x() = A sin(2 π freq ), x2(): symmerical square-wave of frequency freq 2 0 y( ) A. sin 2. π. freq.. A. 2 n =. 2n. sin 2. π. freq. 2 ( 2n. ). A = A 2 = freq = 4 0 3 freq 2 = 3.8 0 4 y() 0 0 5 0 5 0 4.5 0 4 2 0 4 2.5 0 4 3 0 4 3.5 0 4 4 0 4 4.5 0 4 5 0 4 Time domain analysis: Reminder: sin(a) sin (b) = 0.5 [-cos(a+b) + cos(a-b)] Frequency componens of y(): PicoScope Specrum (4096 samples, linear scales) db 38 khz ± 4 khz relaive ampliude: 3 38 khz ± 4 khz /3 5 38 khz ± 4 khz /5 m 38 khz± 4 khz /m 0 0-0 -20-30 -40-50 -60-70 0 50 00 50 200 250 300 khz ADSP/Lab 3. 5 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Frequency domain analysis (convoluion): Concep: X (ω) -3-2 - 0 2 3 ω X 2 (ω) 0 2 3 4 5 6 7 8 9 0 ω X (ω) * X 2 (ω) Procedure (wih AMPMIX board) Board exra-gain: 0 db, Band-pass filer: OFF Inpu a 4 khz sine-wave o he AMPMIX board (Inpu). Adjus is ampliude so as o have he following signal a he amplifier oupu (AMPou): V 2.0.6.2 0.8 0.4 0.0-0.4-0.8 -.2 -.6 µs -2.0 0 50 00 50 200 250 300 350 400 450 500 05Nov2003 3:36 Check ha MIXou is as shown on he nex Figure: 0 2 3 4 5 6 7 8 9 0 ω -2.0 0 50 00 50 200 250 300 350 400 450 500 µs 05Nov2003 3:37 Use PicoScope Specrum Analyser (No of specrum bands: 4096,.5 MHz) o compare your display wih he heory. V 2.0.6.2 0.8 0.4 0.0-0.4-0.8 -.2 -.6 ADSP/Lab 3. 6 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Objecives Laboraory #4 Chap 3 Linear Sysem Response: general case Undersand Frequency down-conversion. Verify ha undesired frequencies (e.g. mirror image) can also conver down o base-band. AMPMIX board bloc diagram: Inpu BPF swich MIXou LPFou Amplifier gain 24 db exra gain: 0, 20, 40 db fc low 0 khz fc high 70 khz Band-Pass Filer fo 40 khz Bw 3.4 khz Bw2 500 Hz AMPou Double Balanced Mixer LPF fc 3.3 khz TTLosc Square-wave generaor + freq. div by 2 fmin 30 khz fmax 60 khz SQWgen Power supply: 5 (8 ma) o 8V (20 ma), Inpu impedance: 5 kω (AC coupling) Amplifier: fc low 0 khz, fc high 70 khz gain: 24 db, addiional gain: + 20 db or + 40 db Band-Pass filer: Second order, resonan frequency adjusable around 40 khz. Bandwidh: Wide (3.4 khz) or Narrow ( 500 Hz) Square-Wave generaor: Tunable fmin 30 khz, fmax 60 khz Exernal frequency conrol (TTLosc), SQWgen freq. = TTLosc freq /2 Double balanced mixer: MIXou() = AMPou() SQWgen(), i.e. AMPou muliplied by ± LPF: Three cascaded R-C low-pass filers ADSP/Lab 3. 7 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Procedure (wih AMPMIX board) Par : Inpu: sine-wave - 40 khz - 00 mv peak-o-peak Exra gain: OFF BPF: OFF SQWgen: Local ( 38 khz) Verify: AMPou: clean 40 khz sine-wave (i.e. no sauraion),.6v pp MIXou: mus look as follows: -5 0 50 00 50 200 250 300 350 400 450 500 3 2 0 - -2-3 -4 µs LPFou: clean 2 khz sine-wave, ampliude:? Change he inpu sine-wave frequency in order o deermine he oupu LPF cuoff frequency. Verify he mirror frequency concep (i.e. inpu signal frequency 36 khz). Verify ha if he inpu signal frequency is 2 khz higher (or lower) han a local oscillaor harmonic, hen down-conversion does occur and LPFou is also a 2 khz sine-wave. Wha can you say of LPFou signal ampliude? Par 2: Inpu: FSK (freq L : 30 khz and freq H : 40 khz), 000 bis/s - 00 mv pp Exra gain: OFF BPF: OFF SQWgen: Local ( 38 khz) Verify: AMPou: clean 30 khz and 40 khz sine-wave (i.e. no sauraion) LPFou looks as follows: 0.5.0.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0-2.0 ms Explain why! Slowly decrease freq H. Wha happens?.2 0.8 0.4 0.0-0.4-0.8 -.2 -.6 ADSP/Lab 3. 8 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003
Objecives Laboraory #5 Chap 3 Linear Sysem Response: general case Undersand... Band-Pass Filer rise-ime Prove...... Procedure...... ADSP/Lab 3. 9 e i a j Wesern Swizerland Universiy of Applied Sciences Prof. J.-P. Sandoz, /2003