what's wrong with this code? (active filter PSpice)

[Katerina]

i am trying to design this bandpass filter [output V(6)] with fo=1kHz and BW=100Hz.
I am sure that my design is corect but Pspice doesn't agree and gives me bandwidth of 500Hz
Is there something wrong with my code?

please help

*Amplifier AC Input Voltage 
*DC Voltage Source 
Vcc  7  0  DC  +15V
Vee  4  0  DC  -15V
*Plot Command (Specified Frequency Range To Be Investigated) 
Vin 1 0 AC 1V
.AC DEC 1 1HZ 3000000HZ
*****CIRCUIT****************************
R1 1 2 159k
R2 2 6 159k
R3 22 26 100K
R10 16 22 100K
R4 6 12 15.9k
R5 2 26 15.9k
C1 2 6 10nF
C2 12 16 10nF
X1 0 2 7 4  6 UA741
X2 0 12 7 4 16 UA741
X3 0 22 7 4 26 UA741
*-----------------------------------------------------------------------------
* connections:   non-inverting input
*                | inverting input
*                | | positive power supply
*                | | | negative power supply
*                | | | | output
*                | | | | |
.subckt uA741    1 2 3 4 5
*
  c1   11 12 8.661E-12
  c2    6  7 30.00E-12
  dc    5 53 dy
  de   54  5 dy
  dlp  90 91 dx
  dln  92 90 dx
  dp    4  3 dx
  egnd 99  0 poly(2),(3,0),(4,0) 0 .5 .5
  fb    7 99 poly(5) vb vc ve vlp vln 0 10.61E6 -1E3 1E3 10E6 -10E6
  ga    6  0 11 12 188.5E-6
  gcm   0  6 10 99 5.961E-9
  iee  10  4 dc 15.16E-6
  hlim 90  0 vlim 1K
  q1   11  2 13 qx
  q2   12  1 14 qx
  r2    6  9 100.0E3
  rc1   3 11 5.305E3
  rc2   3 12 5.305E3
  re1  13 10 1.836E3
  re2  14 10 1.836E3
  ree  10 99 13.19E6
  ro1   8  5 50
  ro2   7 99 100
  rp    3  4 18.16E3
  vb    9  0 dc 0
  vc    3 53 dc 1
  ve   54  4 dc 1
  vlim  7  8 dc 0
  vlp  91  0 dc 40
  vln   0 92 dc 40
.model dx D(Is=800.0E-18 Rs=1)
.model dy D(Is=800.00E-18 Rs=1m Cjo=10p)
.model qx NPN(Is=800.0E-18 Bf=93.75)
.ends
 
.OP
.PROBE
.END

Added after 25 minutes:

and this is my circuit diagram

 

[Katerina]

Anybody...
I tried today the circuit in Shematics but I have the same problem.
Has anybody any idea?
I use the following equation sto calculate R4,R5,R1,R2:
fo=1/2πRC where R=R4=R5 and C=C1=C2
Q=R2/R
they are standard for the Biquad filters