General Engineering Level 4 & 5: 2017

Electrical and Electronics Principles

TWO PORT NETWORKS

A customer has asked you to design a ‘mute’ facility for his hi-fi system. The hi-fi incorporates an amplifier that drives a loudspeaker with an impedance of 50W. The required voltage attenuation is a minimum of 30dB.

Apply two-port network model to the solution of practical problems

a) Explain the concept of characteristic impedance, and calculate and state what value of characteristic impedance is required for this application.

b) Calculate and state how much insertion loss would be required to mute the hifi system.

c) Describe how you would measure insertion loss and the relationship between neper and dB.

Answer:

Attenuation is a way to reduce the amplitude of voltage or current signal between two sources. It is independent of frequency and power. It decreases the gain of an amplifier. Attenuator is opposite of an amplifier.

Project
Attenuation ratio α = 30dB = 20log(V_in/V_out ) = 20log(N_α) → Attenuation factor N_α = 10^3/2
= 10√10 = 31.6228

A Hi-fi power amplifier connected to a loudspeaker. Attenuator of -30dB needed in between. The attenuator here could be a resistor in form of potential divider. It would be a passive symmetrical two port T-connected network as below. Z₁ series resistors and Z₂ is a shunt resistor.

Loudspeaker has very high impedance of 50Ω which is very good and that should match to Hi-Fi output impedance, and characteristic impedance Z_o should be 50Ω making the system and input impedance the same, to ensure maximum power transfer.

Characteristic Impedance Z₀, in our case 50Ω. It’s the ratio of E/I inside a transmission line between two sources. Usually between 50-600Ω. Knowing characteristic impedance ensures E/I ratio being constant. The impedance remains constant between the sources. Higher impedance will mean lesser voltage exiting the line than entering it, low impedance is preferred. On a T-connection it is found by using this formula,

Z₀ = √Z₁²+2·Z₁·Z₂ cross checking above figures Z₀ = √46.9347²+2·46.9347·3.16544) = 50Ω

Insertion loss, for every component introduced in a circuit, there will be a loss of voltage and current. This is not an exception in our case too, as we have introduced a new component ‘attenuator’ between two circuit components; a speaker and an amplifier. Insertion loss A_L found by comparing voltage V_L or current I_L before the new component and after V₂ or I₂, across the load. And it is expressed in decibels. It needs to be 30dB in our case to mute the system and can be measured this way which should give the same for our project;

for cross checking,

Decibel dB uses bas10 logarithmic scale but Neper (Np) uses base e, natural logarithmic scale.
eg . A_L = 20logV_L/V₂ dB = log_eV_L/V₂ Np, conversions 1 dB = ln(10)/20 Np and 1Np = 20/ln(10) dB
30dB = 30·ln(10)/20 = 3.45388Np

Design and test symmetrical attenuators against computer models

a) Using bread board and computer simulation software, design, simulate, build and test, a T-section symmetrical attenuator to provide the required attenuation and characteristic impedance.

b) Determine the propagation coefficient (expressed in terms of attenuation, α, and phase change ß)

c) Determine the input impedance of the attenuator using the simulator. If the loudspeaker was replaced by a loudspeaker with an impedance of 4Ω, describe the effect on the input impedance of the attenuator and the voltage output.

Answer:

Testing the circuit in simulation
Circuit connected to a 16V power source and a switch activates the attenuator. Current and Power over the speaker compared before and after activating the switch as well as measuring V_L and I_L.

						α
	Speaker I	Speaker P		I_L	121.21mA	30.0068dB
Before	164.74mA	217.11mW	&	I₂	3.83mA	30.0068dB
After	3.83mA	117.53μW		V_L	6.06V	31.6219x
α dB	32.6653	32.672		V₂	191.64mV	31.6219x

so the circuit indeed attenuates current and power coming to the speaker by a ratio of at least 30dB and voltage by 31.6219x.

Propagation coefficient ϒ
It tells how signal travels along medium by amplitude and phase. α is the real part and phase change β is the imaginary part, expressed in neper.

Since the above circuit has a DC source which has no frequency hence no phase β = 0.

Replacing loudspeaker with 4Ω

To enable maximum power transfer, impedance of two sources should match. If not there will be a reduction of power transferred meaning the loudspeaker will not work as it should. Hi-fi output is already 50Ω and does not match to 4Ω speaker. Effect of this will be;

· Attenuation will be reduced.

· Less loudspeaker impedance will mean more voltage across it, V₂ will increase.

· This mean attenuator output voltage increases

· I₂ also increases

· As a result, there will be more current & power coming to the speaker.

· V_L and I_L will not change.

· This mean attenuator input impedance will not change either

Looking at the simulator with 50Ω speaker circuit impedance was found by dividing,
Z_in = V_L/I_L = 6.05/121.21E-3 = 49.8805Ω = 50Ω or Z_out = V₂/I₂ = 191.64 / 3.83 = 50.0366Ω = 50Ω

Now, with 4Ω speaker and looking at the test results, attenuator input impedance will be same 50Ω and output impedance found by ,
Z_out = V₂/I₂= 28.39E-3 /7.10E-3 = 3.99859 = 4Ω

If we match Hi-fi output impedance to be 4Ω and loudspeaker 4Ω as well, but attenuator designed for 50Ω input and output. Effect of this will be same as above but a lot higher numbers, and α and impedances will be similar as above. Changing resistor values in the experiment we get these values,

Z_in = V_L/I_L = 9.30/186.24E-3 = 49.9356Ω = 50Ω

Z_out = V₂/I₂= 43.6E-3 /10.90E-3 = 4Ω

This tell that, this attenuator input impedance will always be 50Ω and its output impedance will always be speaker’s impedance, under load.

Practical testing

Attenuation of ~30db as expected

Lab set-up

Attenuation circuit in detail

Electronic engineers will often design, build and test two port networks for various application to include amplifiers, coaxial cables and T- Junction circuits. Study the following two port applications and solve the complex problems

Complex problems involving two port networks with more that one variable

a) If the T – section of a transmission line has a characteristic impedance, Z₀= 400|30⁰W & a short circuit impedance, Z_sc = 600|45⁰W.

Determine the open circuit impedance, Z_oc for the T – Section

Answer:

Z_OC = open circuit impedance Z_SC = short circuit impedance

Z_O =400∠30°Ω Z_SC =600∠45°Ω Z_OC = ?

Z_O = √Z_OC·Z_SC

→ 400∠30° = √Z_OC·600∠45° → 400∠30° · 400∠30° = Z_OC·600∠45° →

= Z_OC

Z_OC= 8E4+j138564/424.26+j424.26 = 257.58+j69.0184 = 266.67∠15°Ω

b) If a T – section composed of pure resistances has Z_oc = 800|0^o
& Z_sc = 600|0⁰.Determine Z₁ & Z₂ for the T – Section.

Answer:

Z_OC =800∠0°Ω Z_SC =600∠0°Ω Z₁ = ? Z₂ = ?

Z_O = √Z_OC·Z_SC= √800∠0°·600∠0° = 480000+j0.00 = √480000.00∠0.00° = 692.82∠0°Ω

Z_OC = 0.5Z₁ +Z₂ → 800 = 0.5Z₁ +Z₂→ 0.5Z₁ = +Z₂-800 → 0.5Z₁ /0.5= (Z₂/0.5)-(800/0.5)

→ Z₁ =1600-(Z₂/0.5)

→692.82²=(1600-(Z₂/0.5))²0.25+(1600-(Z₂/0.5))Z₂ →692.82²=(0.25·400²-(Z₂/0.5)²0.25+ Z₂400-(Z₂Z₂/0.5) →480000= 4000-Z₂²+ 400Z₂-2⁴Z₂→ 480000= 3Z₂400
→ Z₂= 400Ω

Z₁ =1600-(Z₂/0.5) = 1600-(400/0.5) = 800Ω

Z₁ values are 800∠0°Ω and Z₂ value is 400∠0°Ω

Task 3 convergent, lateral, creative thinking are employed to solve problems

You have assessed that the customer also requires a long cable to connect his hifi system in his hallway for an audio PA system. You need to draw up a T-pad attenuator circuit where with the following values:

Z₁ = 200W & Z₂ = 400W.

load is equal to Z₂

a) Calculate the characterstic impedance of the circuit

b) Determinate what will be its insertion loss in dB?

Using the same techniques but in with a variation of requirement, determine: