4 An application of difference equations – currents in a ladder network

The application we will consider is that of finding the electric currents in each loop of the ladder resistance network shown, which consists of ( N + 1 ) loops. The currents form a sequence { i 0 , i 1 , … i N }

Figure 7

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All the resistors have the same resistance R so loops 1 to N are identical. The zero’th loop contains an applied voltage V . In this zero’th loop, Kirchhoff’s voltage law gives

V = R i 0 + R ( i 0 − i 1 )

from which

i 1 = 2 i 0 − V R (23)

Similarly, applying the Kirchhoff law to the ( n + 1 ) t h loop where there is no voltage source and 3 resistors

0 = R i n + 1 + R ( i n + 1 − i n + 2 ) + R ( i n + 1 − i n )

from which

i n + 2 − 3 i n + 1 + i n = 0 n = 0 , 1 , 2 , … ( N − 2 ) (24)

(24) is the basic difference equation that has to be solved.

Task!

Using the left shift theorems obtain the z-transform of equation (24). Denote by I ( z ) the z-transform of { i n } . Simplify the algebraic equation you obtain.

We obtain

z 2 I ( z ) − z 2 i 0 − z i 1 − 3 ( z I ( z ) − z i 0 ) + I ( z ) = 0

Simplifying

( z 2 − 3 z + 1 ) I ( z ) = z 2 i 0 + z i 1 − 3 z i 0 (25)

If we now eliminate i 1 using (23), the right-hand side of (25) becomes

z 2 i 0 + z 2 i 0 − V R − 3 z i 0 = z 2 i 0 − z i 0 − z V R = i 0 z 2 − z − z V i 0 R

Hence from (25)

I ( z ) = i 0 z 2 − 1 + V i 0 R z z 2 − 3 z + 1 (26)

Our final task is to find the inverse z-transform of (26).

Task!

Look at the table of z-transforms on page 35 (or at the back of the Workbook) and suggest what sequences are likely to arise by inverting I ( z ) as given in (26).

The most likely candidates are hyperbolic sequences because both { cosh α n } and { sinh α n } have z-transforms with denominator

z 2 − 2 z cosh α + 1

which is of the same form as the denominator of (26), remembering that cosh α ≥ 1 . (Why are the trigonometric sequences { cos ω n } and { sin ω n } not plausible here?)

To proceed, we introduce a quantity α such that α is the positive solution of 2 cosh α = 3 from which (using cosh 2 α − sinh 2 α ≡ 1 ) we get

sinh α = 9 4 − 1 = 5 2

Hence (26) can be written

I ( z ) = i 0 z 2 − 1 + V i 0 R z z 2 − 2 z cosh α + 1 (27)

To further progress, bearing in mind the z-transforms of { cosh α n } and { sinh α n } , we must subtract and add z cosh α to the numerator of (27), where cosh α = 3 2 .

I ( z ) = i 0 z 2 − z cosh α + 3 z 2 − 1 + V i 0 R z z 2 − 2 z cosh α + 1 = i 0 z 2 − z cosh α z 2 − 2 z cosh α + 1 + 3 2 − 1 z − V z i 0 R z 2 − 2 z cosh α + 1

The first term in the square bracket is the z-transform of { cosh α n } .

The second term is

1 2 − V i 0 R z z 2 − 2 z cosh α + 1 = 1 2 − V i 0 R 2 5 z 5 2 z 2 − 2 z cosh α + 1

which has inverse z-transform

1 2 − V i 0 R 2 5 sinh α n

Hence we have for the loop currents

i n = i 0 cosh ( α n ) + i 0 2 − V R 2 5 sinh ( α n ) n = 0 , 1 , … N (27)

where cosh α = 3 2 determines the value of α .

Finally, by Kirchhoff’s law applied to the rightmost loop

3 i N = i N − 1

from which, with (27), we could determine the value of i 0 .

Exercises
  1. Deduce the inverse z-transform of each of the following functions:
    1. 2 z 2 − 3 z z 2 − 3 z − 4
    2. 2 z 2 + z ( z − 1 ) 2
    3. 2 z 2 − z 2 z 2 − 2 z + 2
    4. 3 z 2 + 5 z 4
  2. Use z-transforms to solve each of the following difference equations:
    1. y n + 1 − 3 y n = 4 n y 0 = 0
    2. y n − 3 y n − 1 = 6 y − 1 = 4
    3. y n − 2 y n − 1 = n y − 1 = 0
    4. y n + 1 − 5 y n = 5 n + 1 y 0 = 0
    5. y n + 1 + 3 y n = 4 δ n − 2 y 0 = 2
    6. y n − 7 y n − 1 + 10 y n − 2 = 0 y − 1 = 16 , y − 2 = 5
    7. y n − 6 y n − 1 + 9 y n − 2 = 0 y − 1 = 1 , y − 2 = 0
    1. ( − 1 ) n + 4 n
    2. 2 + 3 n
    3. cos ( n π ∕ 3 )
    4. 3 δ n − 2 + 5 δ n − 4
    1. y n = 4 n − 3 n
    2. y n = 21 × 3 n − 3
    3. y n = 2 × 2 n − 2 − n
    4. y n = n 5 n
    5. y n = 2 × ( − 3 ) n + 4 × ( − 3 ) n − 3 u n − 2
    6. y n = 12 × 2 n + 50 × 5 n
    7. y n = ( 6 + 3 n ) 3 n