Today we are going to find x of k for the given question x of n 1 2 3 4 Today we are going to find x of k for the given question x of n 1 2 3 4 4 3 2 1 using 4 3 2 1 using DITFFT algorithm. DITFFT algorithm. There are two methods DITFFT and There are two methods DITFFT and DIFFFT. DIFFFT.
We are going to deal with DITFFT. We are going to deal with DITFFT. We know this is X of 0, We know this is X of 0, X of 1, X of 1, X of 2, X of 2, X of 3, X of 3, X of 4, X of 4, X of 5, X of 5, X of 6 and X of X of 6 and X of 7. First of all we need to do bit reversal for 7. First of all we need to do bit reversal for DIT FFT.
DIT FFT. Bit reversal means Bit reversal means 0 0 0 we are writing as 0 0 0 itself. 0 0 0 we are writing as 0 0 0 itself. 0 0 1 we need to write in the reverse order that is 0 0 1 we need to write in the reverse order that is 1 0 0. 1 0 0. Like that 0 0. Like that 0 0. So, So the reverse order will be 010 itself, the reverse order will be 0 1 0 itself.
0 1 1, 011 the reverse order will be 110, the reverse order will be 1 1 0. Here, here 001, 0 0 1, 101 and 011 that is 111. 1 0 1 and 0 1 1, that is triple 1. So this is x of 0 itself. So, this is x of 0 itself. This is x of This is x of 4. 4. Then 0, Then, 0 1, 1. That is x of that is x of 2. 4 plus 2, 2. 4 plus 2, 6. 6. That is x of That is x of 6. x of 6. x of 1. 1 then 4 plus 1 5 x of then 4 plus 1 5 x of 5 5 2 plus 1 3 x of 2 plus 1 3 x of 3 and this x of 7 we need to start with this order for 3 and this x of 7 we need to start with this order for d i d f f d so we know x of 0 it is 1 x of 4 x of 4 it is 4 DIT FFT.
So we know X of 0 it is 1, X of 4 it is 4, then X of 2, then x of 2 x of X of 2 it is 3, 2 it is 3 x of 6 x of 6 it is 2 x of 1 x of 1 it is 2 x of 5 x of 5 it is 3 x of 3 x of 3 it is 4 and X of 6, X of 6 it is 2, X of 1, X of 1 it is 2, X of 5, X of 5 it is 3, X of 3, X of 3 it is 4 and X of 7 it is x of 7 it is 1. we have to start with these terms for drawing the butterfly diagram in d80 fft 1. We have this. start with these terms for drawing the butterfly diagram in DIT FFT. So this is the starting part now we need to put draw these lines after drawing the line we need to put cross marks then after one gap again put cross marks after so this is the starting part now we need to put draw these lines after drawing the line we need to put cross marks then after one gap again put cos marks after a gap put cos mark again after a gap put cos mark then add or put minus one in bottom of all the cos a gap put again after a gap put cos mark then add or put minus 1 in bottom of all the cos marks like this now we need to add the terms we need to find the answer marks like this now we need to add the terms we need to find the answer coming here so first of all coming here first of all look at this line the value in this line is 1 so 1 Look at this line. The value in this line is 1. So, 1, 1, then this line, Then this line 1 plus 4. 1 plus 4. 1 plus 4 it is 1 plus 4 it is 5. 5. Then the second line 4 into minus 1. Then the second line 4 into minus 1 it is minus 4 minus 4 plus 1 minus 4 plus 1 it is minus 3. It is minus 4. Minus 4 plus 1. Minus 4 plus 1 it is. minus 3. Now this line 3, Now this line 3 3 and 3 and 2. 3 plus 2 it is 2, 3 plus 2 it is 5. 5. Now 2 into minus 1 you will get minus 2, Now 2 into minus 1 you will get minus 2, minus 2 plus 3, minus 2 plus 3. 3 minus 2 plus 3 it is minus 2 plus 3 it is 1 now 1. Now 2 plus 3 it is 2 plus 3 it is 5 then 3 into minus 1 minus 3 plus 5. Then 3 into minus 1, minus 3 plus 2 you will get minus 1 now it is 4 plus 1 5 2 you will get minus 1. Now it is 4 plus 1, 5. Then 1 into minus 1, Then 1 into minus 1 we will get minus 1. we will get minus 1. Minus 1 plus 4, Minus 1 plus 4 it is it is 3. So, 3. So we will get the values, we will get the values, the starting value for the next stage. the starting value for the next stage.
We need to multiply with this minus 1, We need to multiply with this minus 1 the second value. the second value. Now we need to extend the line for the next stage.
Now, we need to extend the line for the next stage. We need to extend the line. We need to extend the line.
then Then we put cause marks. We put course marks. We need to put course marks after starting from here after one line put course mark then starting from here after one line put course mark. We need to put cause marks after starting from here after one line put cause mark. Then starting from here after one line put cause mark.
cross mark then do the reverse process starting from this line after one line put cross mark then starting from here after one line join the line this is the first tip or the first section Then do the reverse process starting from this line after one line put course mark. then starting from here after one line join the line this is the first tip of the first section like that we need to do or draw the drawings in the second section also so like that we need to do or draw the drawings in the second section also So starting from here after one line join starting from here after one line join Starting from here after one line join Starting from here after one line join Starting from here after one line join Starting from here after one line again join. Starting from here after one line join Starting from here after one line again join. So we divided the eight lines into two sections So we divided the eight lines into two sections Now we need to put minus one to the bottom of bottom two lines here also for the bottom two lines We need to put minus one Now we need to put minus one to the bottom of bottom two lines here also for the bottom two lines We need to put minus one Now we need to know about weightage.
Now we need to know about weightage. So here the weightage was So here the weightage was W2 raised to 0 that is W2 raised to 0 that is 1. 1. We need to put weightage ahead of the line in DIT F50. We need to put weightage ahead of the line in DIT FFT. For DIT F50, we need to put weightage ahead of the line in DIT F50. For DIT FFT we need to put weightage after the lines, FFT we need to put weightage after the lines after the cross marks so this is only joining two lines so W 2 raise to 0 the weightage will be 1 you need to put weightage here in all these steps after the cross marks.
So this is only joining two lines. So W2 raised to 0 the weightage will be 1. We need to put weightage here. in all these steps i forgot to put that then we need to multiply for this section or the second stage the weightage will be w I forgot to put that then we need to multiply for this section or the second stage the weightage will be w 4 raise to 0 it is 1 and w 4 raise to 1 it is minus j so we divided the 8 lines into two sections and for the first section the last two lines we have to put the weightage 4 raise to 0 it is 1 and w 4 raise to 1 it is minus j so we divided the eight lines into two sections and for the First section, the last two lines, we have to put the weightage.
We are going to put weightage as 1. We are going to put weightage as 1, W 4 raised to 0 is 1 and W 4 raised to minus 1. w4 raise to 0 it is 1 and w4 raise to 0 it is 2. For rest 1 it is minus j. 4 raised to 1 is minus 3. The last two lines we have to put weightage. The last two lines, we have to put weightage.
Here also for the last two lines we have to put the weightage. Here also for the last two lines, we have to put the weightage. 1 and minus 3. 1 and minus j.
Now we are going to do the addition. Now we are going to do the addition. So, So 5 plus 5 it is. 5 plus 5 it is 10. Then, then minus 3, Then minus 3, here you see 1 into minus j.
here you see 1 into minus j. So, So it will be minus j. it will be minus j. So minus 3 minus j. So, minus 3 minus j.
Now, Now 5 into 1 it is 5 into 1, it is 5 into minus 1. 5 into minus 1. See here you can see 1 minus 1. See, here we can see 1 minus 1. So, So 5 into minus 1 it is minus 5. 5 into minus 1, it is minus 5. Minus 5 and plus 5 it will be 0. Minus 5 and plus 5, it will be 0. So here it is 1 into minus j. So here it is 1 into minus j. So minus j into minus 1 it is j. So minus j into minus 1 it is j.
j and minus 3 is coming. j and minus 3 is coming. So minus 3 plus j. So minus 3 plus j.
Minus 3 plus j. Minus 3 plus j. Now in the second section Now in the second section, 5 and it is 5. 5 and it is 5. So 5 plus 5 it will be 10. So 5 plus 5 it will be 10 and minus 1, And minus 1. 3 into minus j, 3 into minus 3. it is minus 3j. minus 3j so minus 1 minus So, minus 1. minus 3j here it will be 5 into minus 1 see 5 into minus 1 it is minus 5 and this one 3j here it will be 5 into minus 1 see 5 into minus 1 it is minus 5 and this one 5 so minus 5 plus 5 it is 0 5 so minus 5 plus 5 it is 0 and 3 into minus 3 it is minus 3j minus 3j into minus 1 it is plus 3j plus 3j and this one it is minus 1 so minus 1 plus 3 into minus 3 is minus 3j minus 3j into minus 1 it is plus 3j plus 3j and this one it is minus 1 so minus 1 plus 3j so this is the starting value for the next stage 3j so this is the starting value for the next stage. Now we need to extend this line for the next stage and that is the final stage.
Now we need to extend this line for the next stage and that is the final stage. Here we need to start from the first line and after 4 lines, Here we need to start from the first line and after 4 lines, the 4th line we need to join. the 4th line we need to join. 1, 1, 2, 2, 3 and 3 and 4. 4. Starting from here after the 4th line. Starting from here after the 4th line.
1, 1, 2, 2, 3 and 4. 3 and 4. line we need to join. line we need to join starting on second line after four lines or on the fourth line we need to join so one two three and four starting from here one two three and four Starting on second line after four lines or on the fourth line we need to join. So one, two, three and four. Starting from here one, two, three and four.
starting from here one two three and four so do the reverse process also starting from here one two three and four here one two three and four one Starting from here, 1, 2, 3 and 4. So do the reverse process also. Starting from here, 1, 2, 3 and 4. Starting from here, 1, 2, 3 and 4. 1, two three and four one two three and four now this is the final stage here we need to put minus one for four lines 2, 3 and 4 1, 2, 3 and 4 Now this is the final stage. Here we need to put minus 1 for 4 lines. The last 4 lines we need to put minus 1. The last four lines we need to put minus one here we put minus one for the last two lines here we put minus one for all lines so now the second thing we need to notice is that we have to put weight teach Here we put minus 1 for the last 2 lines.
Here we put minus 1 for all lines. So now the second thing we need to notice is that we need to put weightage. For the first line, For the first line, the weightage was w2 raised to 0, the weightage was w2 raised to 0, that was 1. that was 1. For the second stage, For the second stage, the weightage was w4 raised to 0, the weightage was w4 raised to 0, it was 1. it was 1. And w4 raised to 1, And w4 raised to 1, it was minus j.
it was minus j. Now, Now, for here, for here, the weightages are w8 raised to 0, the weightages are w8 raised to 0, that is 1. that is 1. W8 raised to 1 that is 0.707 minus 0.707 j and W8 raised to 2 it is minus j and W8 raised to 3 it is minus W 8 raised to 1 that is 0.707 minus 0.707 j and W 8 raised to 2 it is minus j and W 8 raised to 3 it is minus 0.707 minus 0.707 j. 0.707 minus 0.707 j. Now we need to put the weightage for the last four lines. Now we need to put the weightage for the last 4 lines.
1, 1, 2, 2, 3 and 4. 3 and 4. Last 4 lines we put the weightage. Last four lines we put the weightage. So the worst weightage is W8 raise to 0 that is So the worst weightage is W8 raise to 0 that is 1 and W8 raise to 1 that is 0.707 minus 0.707J that is W8 raise to 1. 1 and W8 raise to 1 that is 0.707.
minus 0.707 j that is w8 raised to 1 then w8 raised to 2 that is minus j then w8 raised to 3 it is minus Then, w8 raised to 2, that is minus j. Then, w8 raised to 3, it is minus 0.707, minus 0.707 j. Now we need to multiply the terms. Now, we need to multiply the terms.
First of all we need to add this line. First of all, we need to add this line. So, So 10 plus 10 plus 10 that will give you 20 minus 3 minus j.
10, that will give you 20. minus 3 minus j see here we can see we need to multiply first that is we need to multiply the term the term here it is minus 1 minus 3 j into w See here we can see we need to multiply first. That is we need to multiply the term. The term here it is minus 1 minus 3j into W8 raised to 1. 8 raised to 1 the weightage is w 8 raised to 1 so w 8 raised to 1 it is 0.707 minus 0.707 j The weightage is W8 raised to 1. So W8 raised to 1 it is 0.707 minus 0.707 j. multiplying we get minus 2.828 minus Multiplying we get minus 2.828 minus 1.414 j. 1.414 j so we will get the value as minus 2.828 minus 1.414 j by multiplying these two times and we need to add minus 3 minus j to it we need to add minus 3 minus j to it we So we will get the value as minus 2.828 minus 1.414 j by multiplying this is 2 times and we need to add minus 3 minus j to it.
We need to add minus 3 minus j to it. We will get will get 3 plus 2 to it is 3 plus 2 it is 5.828 minus 2.414 j. 5.828 minus 2.414 j. So, So we will get the answer as 5.828 minus we will get the answer as 5.828 minus 2.414 j as the second time.
2.414 j as the second step. Like that do all the same. Like that do all the process. the process so here it will be So, here it will be 0 and 0 and 0 the third term will be 0 and the fourth term it will be minus 3 plus j and we need to multiply here see minus 1 plus 3 j and w 8 raise to 0. The third time will be 0. and the fourth term it will be minus 3 plus j and we need to multiply here see minus 1 plus 3 j and w 8 raised to 3 minus 1 plus 3 j and w 8 raised to 3 it is 0.707 minus 0.70 j so we will get the final answer here so minus 1 plus 3j and w it raise to 3 it is 0.707 minus 0.70j so we will get the final answer here so we will get the answer like this so this will be our x of we will get the answer like this so this will be our x of k the first value will be x of The first value will be x of 0, this is our x of 1, 0 and this is our x of 1. x of 2, x of 2, x of 3, x of 3, x of 4, x of 4, x of 5, x of 5, x of 6 and x of x of 6 and x of 7. 7. if I inline x of k equal to Final answer x of k equal to 20, 20 comma. 5.828 minus 2.414j, 5.828 minus 2.414 j, 0, minus 0.172 minus j.414j, 0 minus 0.172 minus j.414 j, 0, minus 0.172 plus 0 minus 0.172 plus 0.414 j, 0.414j, 0. 0. and minus 5.82 plus j 2.414 so this will be our answer and minus 5.82 plus j 2.414 so this will be our answer