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Home Work 10
The problems in this problem set cover lectures C1 1 and C1 2
a. Define a recursive binary search algorithm.
If lb > ub Return - else Mid := (lb+ub)/ If Array(Mid) = element Return Mid Elsif Array(Mid) < Element Return Binary_Search(Array, mid+1, ub, Element) Else Return Binary_Search(Array, lb, mid-1, Element) End if End if
b. Implement your algorithm as an Ada95 program.
- function Binary_Search (My_Search_Array : My_Array; Lb : Integer; Ub: Integer; Element : Integer) return Integer is
- mid : integer;
- begin
- if (Lb> Ub) then
- return -1;
- else
- Mid := (Ub+Lb)/2;
- if My_Search_Array(Mid) = Element then
- return(Mid);
- elsif My_Search_Array(Mid) < Element then
- return (Binary_Search(My_Search_Array, Mid+1, Ub, Element));
- else
- return (Binary_Search(My_Search_Array, Lb, Mid-1, Element));
- end if;
- end if;
- end Binary_Search;
- end Recursive_Binary_Search;
c. What is the recurrence equation that represents the computation time of your algorithm?
Recursive Binary Search Cost if (Lb> Ub) then c return -1; c else c Mid := (Ub+Lb)/2; c if My_Search_Array(Mid) = Element then c return(Mid); c elsif My_Search_Array(Mid) < Element then c return (Binary_Search(My_Search_Array, Mid+1, Ub, Element)); T(n/2) else c return (Binary_Search(My_Search_Array, Lb, Mid-1, Element)); T(n/2) end if; c end if; c
In this case, only one of the recursive calls is made, hence only one of the T(n/2) terms is included in the final cost computation.
Therefore T(n) = (c1+c2+c3+c4+c5+c6+c7+c8+c9+c10) + T(n/2) = T(n/2) + C
d. What is the Big-O complexity of your algorithm? Show all the steps in the computation based on your algorithm.
T(n) = T(n/2) + C ¥ T(n) = aT(n/b) + cnk , where a,c > 0 and b > 1
T(n) =
log (^) b a
O?? a? b
k
n
k k
O? n log bn? a? b
k^ 1 = 2^0 , hence the second term is used,
??? b
k
T(n) =
- What is the Big-O complexity of : a. Heapify function
A heap is an array that satisfies the heap properties i.e., A(i) ≤ A(2i) and A(i) ≤ A(2i+1).
Simplifying => T(n) = O(n log(n) )
c. Heap_Sort
Heap Sort Cost t(n)
Build_Heap(Heap_Array, Size); O(nlogn)) for I in reverse 2.. size loop n Swap(Heap_Array, 1, I); c1(n-1) Heap_Size:= Heap_Size -1; c2(n-1) Heapify(Heap_Array, 1); O(log n)(n-1)
T(n) = 2 O(nlogn) + (c1+c2+1)n - O(log n) + = 2 O(nlog n) - O(log n) + c‘n
Simplifying, => T(n) = O(nlogn)
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Unified Engineering II Spring 2004
Problem S10 (Signals and Systems) Solution
- Because the numerator is the same order as the denominator, the partial fraction expansion will have a constant term:
3 s^2 + 3s − 10 G(s) = (^2) s − 4 3 s^2 + 3s − 10 = (s − 2)(s + 2) b c = a + + s − 2 s + 2
To find a, b, and c, use coverup method:
a = G(s)| (^) s=∞= 3 3 s^2 + 3s − 10 b = s + 2
s= 3 s^2 + 3s − 10 c = = 1 s − (^2) s=− 2
So 2 1 G(s) = 3 + s − 2
s + 2
, Re[s] > 2
We can take the inverse LT by simple pattern matching. The result is that
g(t) = 3δ(t) + 2 e^2 t^ + e−^2 t^ σ(t)
6 s^2 + 26s + 26 G(s) = (s + 1)(s + 2)(s + 3) a b c = + + s + 1 s + 2 s + 3 Using partial fraction expansions,
6 s^2 + 26s + 26 (s + 2)(s + 3)
a = = 3 s=− 1 6 s^2 + 26s + 26 b = = 2 (s + 1)(s + 3) (^) s=− 2 6 s^2 + 26s + 26 c = = 1 (s + 1)(s + 2) (^) s=− 3
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So 4 s^3 + 11s^2 + 5s + 2 a 2 c 4 G(s) = = + + + s^2 (s + 1)^2 s s^2 s + 1 (s + 1)^2 To find a and c, pick two values of s, say, s = 1 and s = 2. Then 4 + 11 + 5 + 2 a 2 c 4 G(1) = = + + + 12 (1 + 1)^2 1 12 1 + 1 (1 + 1)^2 4 23 + 11 22 + 5 2 + 2 a 2 c 4 G(2) =
22 (2 + 1)^2 2 22 2 + 1 (2 + 1)^2
Simplifying, we have that c 5 a + = 2 2 a c 3
- = 2 3 2 Solving for a and c, we have that
a = 1 c = 3
So 1 2 3 4 G(s) = + + + s s^2 s + 1 (s + 1)^2 and g(t) = 1 + 2t + 3e−t^ + 4te−t^ σ(t)
- G(s) can be expanded as
s^3 + 3s^2 + 9s + 12 G(s) = (s^2 + 4) (s^2 + 9) s^3 + 3s^2 + 9s + 12 = (s + 2j)(s − 2 j)(s + 3j)(s − 3 j) a b c d = + + + s + 2j s − 2 j s + 3j s − 3 j The coefficients can be found by the coverup method: s^3 + 3s^2 + 9s + 12 (s − 2 j)(s + 3j)(s − 3 j)
a = = 0. 5 s=− 2 j s^3 + 3s^2 + 9s + 12 b = = 0. 5 (s + 2j)(s + 3j)(s − 3 j) (^) s=+2j s^3 + 3s^2 + 9s + 12 c = = 0. 5 j (s + 2j)(s − 2 j)(s − 3 j) s^3 + 3s^2 + 9s + 12 d = (s + 2j)(s − 2 j)(s + 3j)
s=− 3 j
s=+3j
= − 0. 5 j
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Therefore
5 0. 5 G(s) = + +
5 j
− 0. 5 j , Re[s] > 0 s + 2j s − 2 j s + 3j s − 3 j
and the inverse LT is
g(t) = 0. 5 e−^2 jt^ + e^2 jt^ + je−^3 jt^ − je^3 jt σ(t)
This can be expanded using Euler’s formula, which states that
eajt^ = cos at + j sin at
Applying Euler’s formula yields
g(t) = (cos 2t + sin 2t) σ(t)
(^00 100 200 300 400 500 )
20
40
60
80
100
120
140
160
180
Impulse response, g(t)
Time, t (sec)
- From the problem statement,
H(s) k G¯(s)
R(s) 1 + k G¯(s) 1 k (^2) s(s + 2ζωns + ω^2 n) = (^1) 1 + k s(s^2 + 2ζωns + ω^2 n) k = s^3 + 2ζωns^2 + ω^2 n^ s + k
So the poles of the system are the roots of the denominator polynomial,
φ(s) = s^3 + 2ζωns^2 + ω^2 s + k = 0 n
The roots can be found using Matlab, a programmable calculator, etc. The plot of the roots (the “root locus”) is shown below. Note that the oscillatory poles go unstable at a gain of only k = 0.000118.
-0.5-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.
-0.
-0.
-0.
-0.
0
Real part of s
Imaginary part of s
- The roots locus for negative gains can be plotted in a similar way, as below. Note that the real pole is unstable for all negative k.
-0.5-0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.
-0.
-0.
-0.
-0.
0
Real part of s
Imaginary part of s
the integral converges only for Re[s] < a. To find the integral, integrate by parts: � (^0) G(s) = te(a−s)t^ dt −∞ 1 a − s
e(a−s)t
0 1 �^0
= t − e(a−s)t^ dt a − s (^) −∞ � (^0) −∞ (^1) (a−s)t = 0 − e dt a − s (^) −∞ 1 0 = − e (a − s)^2
(a−s)t −∞ 1 = − Re[s] < a (s − a)^2
g(t) = cos(ω t 0 t)^ e −a| |, for all t
The LT is given by ∞ cos(ω^ t 0 t)^ e G(s) = −a| |e−st^ dt −∞ For the LT to converge, the integrand must go to zero as t goes to −∞ and ∞. Therefore, the integral converges only for −a < Re[s] < a. The integral is given by ∞ G(s) = cos(ω 0 t) e−a|t|e−st^ dt −∞ � (^0) ∞ at (^) e−st (^) dt cos(ω 0 t)^ e cos(ω −at^ e−st^ dt = 0 t) e + −∞ 0 Expanding the cosine term as
ejω^0 t^ + e−jω^0 t cos(ω 0 t) = 2 yields � (^0) ejω^0 t^ + e−jω^0 t^ at ∞ejω^0 t^ + e−jω^0 t G(s) = e e−st^ dt + e−at^ e−st^ dt −∞^2 � (^0) e(jω^0 +a−s)t^ + e(−jω^0 +a−s)t dt +
∞e(jω 0 −a−s)t (^) + e(−jω 0 −a−s)t = dt −∞^2 0 1 1 1 1 1 = + 2 jω 0 + a − s −jω 0 + a − s
jω 0 − a − s
−jω 0 − a − s s + a =
s − a , −a < Re[s] < a s^2 + 2as + a^2 + ω^2
0 s^2 −^2 as^ +^ a^2 +^ ω^20
