LTI System Review and Equivalent Circuits
Samantha R Summerson
8 September, 2009
1
LTI Systems: a (brief ) Review
A system is linear if it has the following two properties:
1. Superposition: If 𝑥1 (𝑡) → 𝑦1 (𝑡) and 𝑥2 (𝑡) → 𝑦2 (𝑡), then
𝑥1 (𝑡) + 𝑥2 (𝑡) → 𝑦1 (𝑡) + 𝑦2 (𝑡).
2. Scaling: If 𝑥(𝑡) → 𝑦(𝑡), then for a constant 𝑎,
𝑎𝑥(𝑡) → 𝑎𝑦(𝑡).
A system is time invariant if for any 𝜏 ∈ ℝ,
𝑥(𝑡 − 𝜏 ) → 𝑦(𝑡 − 𝜏 ).
If a system that is both linear and time-invariant, we call it a LTI system. Note that the properties are
independent of each other - one may have a linear time-varying system or a non-linear time invariant system.
Example 1. Consider an LTI system with the following input-output pair. What is the output of the
system when the input is 𝑥1 (𝑡) (shown below)?
𝑥(𝑡)
𝑦(𝑡)
1
1
→
1
𝑡
1
𝑡
Figure 1: Input-output pair.
We can find the output by first re-writing 𝑥1 (𝑡) in terms of the input 𝑥(𝑡).
𝑥1 (𝑡) = 3𝑥(𝑡 + 2) + 𝑥(−𝑡) + 𝑥(𝑡) + 3𝑥(𝑡 − 2)
By properties of linearity and time-invariance, we know that the output will be
𝑦1 (𝑡) = 3𝑦(𝑡 + 2) + 𝑦(−𝑡) + 𝑦(𝑡) + 3𝑦(𝑡 − 2).
1
𝑥1 (𝑡)
3
1
1
2
𝑡
Figure 2: Input, 𝑥1 (𝑡).
𝑦1 (𝑡)
3
1
1
2
𝑡
Figure 3: Output, 𝑦1 (𝑡).
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Equivalent Resistance Revisited and Voltage/Current Dividers
In class, we learned equivalent resistance formulas for resistors. If the resistors are in series,
∑
𝑅𝑒𝑞 =
𝑅𝑘 .
𝑘
If the resistors are in parallel,
𝑅𝑒𝑞
(
)−1
∑ 1
=
.
𝑅𝑘
𝑘
Example 2. Find the equivalent resistance for the circuit shown below in Fig 2.
The resistor 𝑅1 is in series with the parallel pair, 𝑅2 and 𝑅3 , and with 𝑅4 . Thus,
𝑅𝑒𝑞 = 𝑅1 + (𝑅2 ∣∣𝑅3 ) + 𝑅4 .
Example 3. Find the equivalent resistance for the circuit shown below in Fig 4. In this example, the second
resistor, 𝑅2 , is short-circuited. Thus,
𝑅𝑒𝑞 = (𝑅1 ∣∣𝑅3 ).
Related to the idea of equivalent resistance are the voltage divider and current divider rules. For two
resistors in series with a voltage source 𝑣𝐼𝑁 , the voltage 𝑣𝑘 across resistor 𝑅𝑘 is
𝑣𝑘 =
𝑅𝑘
𝑣𝐼𝑁 .
𝑅𝑘 + 𝑅𝑗
2
𝑅1
𝑅2
𝑅3
𝑅4
Figure 4: Example - find equivalent resistance.
𝑅2
𝑅1
𝑅3
Figure 5: Example - find equivalent resistance.
Here, 𝑅𝑗 is the resistance of the other resistor. For two resistors in parallel with a current source, 𝑖𝐼𝑁 , the
current through the resistor 𝑅𝑘 is
𝑅𝑗
𝑖𝑘 =
𝑖𝐼𝑁 .
𝑅𝑘 + 𝑅𝑗
Again, 𝑅𝑗 is the resistor of the other resistor (i.e. not the one we’re measuring current through). Note the
difference in the numerator in the two formulas.
Example 4. Find the load resistance values which cause half the current and a quarter of the current to go
through 𝑅𝐿 . First, we want to find 𝑅𝐿 such that 𝑖𝐿 = 12 𝑖𝐼𝑁 . Using our current divider rule, we find
𝑖𝐼𝑁 ↑
𝑅𝐿
𝑅
Figure 6: Example - find values of 𝑅𝐿 .
3
1
2 𝑖𝐼𝑁
=
𝑖𝐼𝑁
⇒
𝑅
,
𝑅 + 𝑅𝐿
1
(𝑅 + 𝑅𝐿 ) = 𝑅,
2
1
1
𝑅,
⇒ 𝑅𝐿 =
2
2
⇒ 𝑅𝐿 = 𝑅.
When the two resistors are of equal value, the current from the source will equally split between them. Now,
we wish to find 𝑅𝐿 such that 𝑖𝐿 = 41 𝑖𝐼𝑁 . Again, we use current divider.
1
4 𝑖𝐼𝑁
=
𝑅
,
𝑅 + 𝑅𝐿
𝑖𝐼𝑁
1
⇒ (𝑅 + 𝑅𝐿 ) = 𝑅,
4
3
1
𝑅,
⇒ 𝑅𝐿 =
4
4
⇒ 𝑅𝐿 = 3𝑅.
We must triple the value of the other resistor in order for one quarter of the current to pass through our
load resistance.
3
Thévenin and Mayer-Norton Circuits
𝑅𝑒𝑞
+
𝑖
𝑣𝑒𝑞
+
±
𝑣
−
−
Figure 7: Thévenin equivalent circuit.
+
𝑖𝑠𝑐
↑
𝑅𝑒𝑞
𝑣
−
Figure 8: Mayer-Norton equivalent circuit.
We like to simplify circuits by finding equivalent resistances for parallel and series resistors; the idea of
simplification can be taken a step further. For any circuit composed of sources and resistors, we can find a
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Thévenin or Mayer-Norton equivalent circuit. A Thévenin equivalent consists of a voltage source in series
with a resistor, whereas a Mayer-Norton equivalent consists of a current source in parallel with a resistor.
How do we find the equivalent voltage, current, and resistance?
1. Consider the open circuit (𝑖 = 0). Find the voltage, 𝑣𝑒𝑞 = 𝑣𝑜𝑐 .
2. Consider the short circuit, setting the terminal voltage to zero (𝑣 = 0). Solve for 𝑖 = 𝑖𝑠𝑐 .
3. Zero the sources (all voltage sources short-circuited and current sources open-circuited) and find the
equivalent resistance, 𝑅𝑒𝑞 , looking back into the terminals.
𝑅𝑒𝑞 = −
𝑣𝑒𝑞
𝑖𝑠𝑐
We can find the equivalent circuit by performing any two of the three above steps. We can solve for
the third unknown quantity using the 𝑣 − 𝑖 relation.
Note that if we short-circuit our Thévenin equivalent circuit, the current traveling clockwise is 𝑖𝑠𝑐 .
Example 5. Find the Norton equivalent circuit. For sake of completeness, we will go through all three
15Ω
𝑖1 𝑖2
𝑖
+
𝑖3
10𝑉 ±
↑ 1𝐴
15Ω
𝑣
−
Figure 9: Example - find the Norton equivalent circuit.
steps.
(a) Let 𝑖 = 0. Using KCL we have
𝑖2 + 1
𝑖1
=
0,
= 𝑖2 + 𝑖3 .
Putting the two together, we get
𝑖1 = −1 + 𝑖3 .
Using KVL, along with the KVL, we have
𝑣
10
=
25𝑖3 ,
=
25(𝑖1 + 1),
=
15𝑖1 + 25𝑖3
=
15𝑖1 + 25(𝑖1 + 1)
=
40𝑖1 + 25.
Solving for 𝑖1 , we get
3
𝑖1 = − .
8
5
Now,
𝑣
=
25(𝑖1 + 1)
3
= 25(− + 1)
8
125
=
8
Thus,
𝑣𝑒𝑞 = 𝑣𝑜𝑐 =
125
𝑉.
8
(b) Set 𝑣 = 0, i.e. put a wire across the terminal. Using KVL, we see that
25𝑖3 = 0.
This means that 𝑖3 = 0, so it is as if an open circuit replaced the resistor of value 25Ω. Thus,
doing KVL for the outer loop,
15𝑖1
=
⇒ 𝑖1
=
10,
2
.
3
Now we can use KCL and plug in this value for 𝑖1 .
𝑖 + 𝑖1 + 1 = 0,
2
⇒ 𝑖 + + 1 = 0,
3
5
⇒𝑖=− .
3
Thus, 𝑖𝑠𝑐 = 𝑖 = − 53 𝐴.
(c) Finally, we consider zero-ing out all the sources. From the terminal’s point of view, the two
resistors are in parallel.
𝑅𝑒𝑞
=
=
=
(15∣∣25),
(
)−1
1
1
+
,
15 25
75
Ω.
8
From the above steps it should be clear that only two are really necessary.
6
15Ω
𝑖1 𝑖2
𝑖
𝑖3
10𝑉 ±
↑ 1𝐴
15Ω
𝑣=0
Figure 10: Example 5 - Step 2 (finding short-circuit current).
75
8 Ω
+
+
125
8 𝑉
±
𝑣
−
−
Figure 11: Example 5 - Thévenin equivalent circuit.
+
5
3𝐴
75
8 Ω
↑
𝑣
−
Figure 12: Example 5 - Mayer-Norton equivalent circuit.
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